NCERT Solutions Class 12 Maths Chapter 7 Integrals Integrals from Chapter 7 of Class 12 introduces students to an important concept called integration, which is the reverse process of differentiation. This chapter explains types of integrals - indefinite and definite, different methods of integration, and how standard integrals are used to solve problems. Integrals are a key part of calculus and are extremely essential for understanding applications like area under the curves, which are covered in later chapters.
Download NCERT Solutions for Class 12 Maths Chapter 7 The NCERT Solutions are prepared strictly as per the latest NCERT syllabus and are fully aligned with the syllabus prescribed by CBSE. To make it easier for the students of Class 12 to understand formulas and methods, the solutions are explained in a step by step manner using simple language. Regular practice helps improve accuracy, builds confidence, and prepares students well for board exams and competitive tests.
1.0 Key Concepts of Class 12 Maths Chapter 7 Integrals Class 12 Maths Chapter 7, Integrals, introduces integration and explains different methods used to find integrals. The main topics covered in NCERT Solutions for Class 12 Maths Chapter 7 include:
Integration as an Inverse Process: Understand integration as the reverse process of differentiation and learn its basic meaning. Indefinite Integrals: Learn standard integrals and solve integrals without fixed limits . Methods of Integration: Study methods such as substitution, integration by parts, and partial fractions to solve different types of integrals. Integrals of Special Functions: Practise integration of trigonometric, exponential, and rational functions. Definite Integrals: Understand definite integrals with limits and learn how to evaluate them. Properties of Definite Integrals: Apply important properties of definite integrals to simplify calculations and solve questions efficiently. 2.0 NCERT Class 12 Maths Chapter 7 Integrals : Detailed Solutions EXERCISE - 7.1
Find an anti-derivative (or integral) of the following function by the method of inspection.
sin 2 x
Sol. The anti derivative of sin 2 x is a function of x whose derivative is sin 2 x .
It is known that,
∵ dx d ( cos 2 x ) = − 2 sin 2 x
⇒ − 2 1 dx d ( cos 2 x ) = sin 2 x
⇒ dx d ( − 2 1 cos 2 x ) = sin 2 x Therefore, the anti-derivative of sin 2 x is − 2 1 cos 2 x
cos 3 x
Sol. The anti derivative of cos 3 x is a function of x whose derivative is cos 3 x .
∵ dx d ( sin 3 x ) = 3 cos 3 x
⇒ 3 1 dx d ( sin 3 x ) = cos 3 x
⇒ dx d ( 3 1 sin 3 x ) = cos 3 x Therefore, the anti derivative of cos 3 x is 3 1 sin 3 x .
e 2 x
Sol. The anti derivative of e 2 x is the function of x whose derivative is e 2 x .
∵ dx d ( e 2 x ) = 2 e 2 x
⇒ 2 1 dx d ( e 2 x ) = e 2 x
⇒ dx d ( 2 1 e 2 x ) = e 2 x Therefore, the anti derivative of e 2 x is 2 1 e 2 x .
( ax + b ) 2
Sol. The anti derivative of ( ax + b ) 2 is the function of x whose derivative is ( ax + b ) 2 .
It is known that,
∵ dx d ( ax + b ) 3 = 3 a ( ax + b ) 2
⇒ 3 a 1 dx d ( ax + b ) 3 = ( ax + b ) 2
⇒ dx d ( 3 a 1 ( ax + b ) 3 ) = ( ax + b ) 2 Therefore, the anti derivative of ( a x + b ) 2 is 3 a 1 ( ax + b ) 3
sin 2 x − 4 e 3 x
Sol. The anti derivative of ( sin 2 x − 4 e 3 x ) is the function of x whose derivative is ( sin 2 x − 4 e 3 x ) .
∵ dx d ( cos 2 x ) = − 2 sin 2 x
⇒ − 2 1 dx d ( cos 2 x ) = sin 2 x
⇒ dx d ( − 2 1 cos 2 x ) = sin 2 x Similarly, dx d ( 3 4 e 3 x ) = 4 e 3 x
∴ dx d ( − 2 1 cos 2 x ) − dx d ( 3 4 e 3 x ) = sin 2 x − 4 e 3 x
⇒ dx d ( − 2 1 cos 2 x − 3 4 e 3 x ) = sin 2 x − 4 e 3 x
Therefore, the anti derivative of ( sin 2 x − 4 e 3 x ) is ( − 2 1 cos 2 x − 3 4 e 3 x )
Find the following integrals in (Q. 6 to 20)
∫ ( 4 e 3 x + 1 ) dx
Sol. Let I = ∫ ( 4 e 3 x + 1 ) dx = 4 ∫ e 3 x dx + ∫ 1 dx = 4 ( 3 e 3 x ) + x + C = 3 4 e 3 x + x + C
∫ x 2 ( 1 − x 2 1 ) d x Sol.
∫ x 2 ( 1 − x 2 1 ) d x = ∫ ( x 2 − 1 ) d x = ∫ x 2 d x − ∫ 1 d x = 3 x 3 − x + C
∫ ( a x 2 + b x + c ) d x Sol.
Let I = ∫ ( a x 2 + b x + c ) d x = a ∫ x 2 d x + b ∫ x d x + c ∫ 1. d x = a ( 3 x 3 ) + b ( 2 x 2 ) + c x + C = 3 a x 3 + 2 b x 2 + c x + C
∫ ( 2 x 2 + e x ) d x Sol.
Let I = ∫ ( 2 x 2 + e x ) d x = 2 ∫ x 2 d x + ∫ e x d x = 2 ( 3 x 3 ) + e x + C = 3 2 x 3 + e x + C
∫ ( x − x 1 ) 2 d x Sol.
Let I = ∫ ( x − x 1 ) 2 d x = ∫ ( x + x 1 − 2 ) d x = ∫ x d x + ∫ x 1 d x − 2 ∫ 1. d x = 2 x 2 + log ∣ x ∣ − 2 x + C
∫ x 2 x 3 + 5 x 2 − 4 d x Sol.
Let I = ∫ x 2 x 3 + 5 x 2 − 4 d x = ∫ ( x + 5 − 4 x − 2 ) d x = ∫ x d x + 5 ∫ 1. d x − 4 ∫ x − 2 d x = 2 x 2 + 5 x − 4 ( − 1 x − 1 ) + C = 2 x 2 + 5 x + x 4 + C
∫ x x 3 + 3 x + 4 dx Sol. Let
I = ∫ x x 3 + 3 x + 4 dx = ∫ ( x 2 5 + 3 x 2 1 + 4 x 2 − 1 ) dx ( ∵ ∫ x n dx = n + 1 x n + 1 + C ) = 2 7 x ( 2 7 ) + 2 3 3 ( x 2 3 ) + 2 1 4 ( x 2 1 ) + C = 7 2 x 2 7 + 2 x 2 3 + 8 x 2 1 + C = 7 2 x 2 7 + 2 x 2 3 + 8 x + C
∫ x − 1 x 3 − x 2 + x − 1 dx Sol. Let I = ∫ x − 1 x 3 − x 2 + x − 1 dx
On dividing, we obtain
I = ∫ x − 1 x 2 ( x − 1 ) + 1 ( x − 1 ) d x = ∫ ( x − 1 ) ( x − 1 ) ( x 2 + 1 ) d x = ∫ x 2 d x + ∫ 1 ⋅ d x = 3 x 3 + x + C
∫ ( 1 − x ) x d x Sol.
Let I = ∫ ( 1 − x ) x d x = ∫ ( x − x 2 3 ) d x = ∫ x 2 1 d x − ∫ x 2 3 d x = 3/2 x 2 3 − 5/2 x 2 5 + C = 3 2 x 3/2 − 5 2 x 5/2 + C
∫ x ( 3 x 2 + 2 x + 3 ) d x Sol.
Let I = ∫ x ( 3 x 2 + 2 x + 3 ) d x = ∫ ( 3 x 2 5 + 2 x 2 3 + 3 x 2 1 ) d x = 3 ∫ x 2 5 d x + 2 ∫ x 2 3 d x + 3 ∫ x 2 1 d x = 3 ( 2 7 x 2 7 ) + 2 ( 2 5 x 2 5 ) + 3 ( 2 3 x 2 3 ) + C = 7 6 x 2 7 + 5 4 x 2 5 + 2 x 2 3 + C
∫ ( 2 x − 3 cos x + e x ) d x Sol.
Let I = ∫ ( 2 x − 3 cos x + e x ) d x = 2 ∫ x d x − 3 ∫ cos x d x + ∫ e x d x = 2 2 x 2 − 3 ( sin x ) + e x + C = x 2 − 3 sin x + e x + C
∫ ( 2 x 2 − 3 sin x + 5 x ) d x Sol.
Let I = ∫ ( 2 x 2 − 3 sin x + 5 x ) d x = 2 ∫ x 2 d x − 3 ∫ sin x d x + 5 ∫ x 2 1 d x = 3 2 x 3 − 3 ( − cos x ) + 5 ( 2 3 x 2 3 ) + C = 3 2 x 3 + 3 cos x + 3 10 x 2 3 + C
∫ sec x ( sec x + tan x ) d x Sol.
Let I = ∫ sec x ( sec x + tan x ) dx = ∫ ( sec 2 x + sec x tan x ) dx = tan x + sec x + C
∫ cosec 2 x s e c 2 x dx
Sol. Let I = ∫ cosec 2 x s e c 2 x dx = ∫ s i n 2 x 1 c o s 2 x 1 dx = ∫ cos 2 x sin 2 x d x = ∫ tan 2 x d x = ∫ ( sec 2 x − 1 ) d x = ∫ sec 2 x d x − ∫ 1 d x = tan x − x + C
∫ c o s 2 x 2 − 3 s i n x d x Sol.
Let I = ∫ cos 2 x 2 − 3 sin x d x = ∫ ( cos 2 x 2 − cos 2 x 3 sin x ) d x = ∫ 2 sec 2 x d x − 3 ∫ tan x sec x d x = 2 tan x − 3 sec x + C
Choose the correct answer in the following Exercises 21 and 22
The anti derivative of ( x + x 1 ) equals?
(A) 3 1 x 3 1 + 2 x 2 1 + C
(B) 3 2 x 3 2 + 2 1 x 2 + C
(C) 3 2 x 2 3 + 2 x 2 1 + C
(D) 2 3 x 2 3 + 2 1 x 2 1 + C
Sol. (C) Let I = ∫ ( x + x 1 ) d x = ∫ x 2 1 d x + ∫ x − 2 1 d x = 3/2 x 2 3 + 1/2 x 2 1 + C = 3 2 x 2 3 + 2 x 2 1 + C
Hence, the correct Answer is C.
mIf dx d ( f ( x )) = 4 x 3 − x 4 3 such that f ( 2 ) = 0 , then f ( x ) is?
(A) x 4 + x 3 1 − 8 129
(B) x 3 + x 4 1 + 8 129
(C) x 4 + x 3 1 + 8 129
(D) x 3 + x 4 1 − 8 129 Sol. (A) It is given that, dx d ( f ( x )) = 4 x 3 − x 4 3
∴ Integrating both sides w.r.t. to x
∴ f ( x ) = ∫ ( 4 x 3 − x 4 3 ) dx
⇒ f ( x ) = 4 ∫ x 3 dx − 3 ∫ ( x − 4 ) dx
⇒ f ( x ) = 4 ( 4 x 4 ) − 3 ( − 3 x − 3 ) + C
∴ f ( x ) = x 4 + x 3 1 + C
Also, f ( 2 ) = 0
∴ f ( 2 ) = ( 2 ) 4 + ( 2 ) 3 1 + C = 0
⇒ 16 + 8 1 + C = 0
⇒ C = − ( 16 + 8 1 )
⇒ C = − 8 129
Put value of C in eq. (1)
f ( x ) = x 4 + x 3 1 − 8 129 .
Hence, the correct Answer is (A).
EXERCISE 7.2 Integrate the functions in exercises 1 to 37
∫ 1 + x 2 2 x dx
Sol. Let I = ∫ 1 + x 2 2 x dx
Put 1 + x 2 = t ⇒ 2 xdx = dt ⇒ I = ∫ t 1 dt = log ∣ t ∣ + C = log 1 + x 2 + C = log ( 1 + x 2 ) + C
∫ x ( l o g ∣ x ∣ ) 2 dx
Sol. Let I = ∫ x ( l o g ∣ x ∣ ) 2 dx
Put log ∣ x ∣ = t ⇒ x 1 dx = dt
⇒ I = ∫ t 2 dt = 3 t 3 + C
= 3 ( l o g ∣ x ∣ ) 3 + C ∫ x + x l o g x 1 dx
Sol. Let I = ∫ x + x l o g x 1 dx = ∫ x ( 1 + l o g x ) 1 dx
Put 1 + log x = t ⇒ x 1 dx = dt
⇒ I = ∫ t 1 dt = log ∣ t ∣ + C = log ∣1 + log x ∣ + C ∫ sin x ⋅ sin ( cos x ) dx
Sol. Let I = ∫ sin x . sin ( cos x ) dx
Put cos x = t ⇒ − sin x dx = dt
⇒ I = − ∫ sin tdt
= − [ − cost ] + C
= cos ( cos x ) + C ∫ sin ( ax + b ) cos ( ax + b ) dx
Sol. Let I = ∫ sin ( ax + b ) cos ( ax + b ) dx
= 2 1 ∫ sin 2 ( ax + b ) dx
Put 2 ( ax + b ) = t ⇒ dx = 2 a dt
⇒ I = 2 1 ∫ 2 a s i n tdt = 4 a 1 [ − cos t ] + C
= 4 a − 1 cos 2 ( ax + b ) + C ∫ ax + b dx
Sol. Let I = ∫ ax + b dx
Put ax + b = t 2 ⇒ adx = 2 tdt
∴ I = ∫ t ⋅ a 2 t d t = a 2 ∫ t 2 d t = a 2 ⋅ 3 t 3 + C = 3 a 2 ( a x + b ) 3/2 + C
∫ x x + 2 d x
Sol. Let I = ∫ x x + 2 dx
Put x + 2 = t 2 ⇒ dx = 2 tdt
∴ I = ∫ ( t 2 − 2 ) . t .2 t dt = ∫ ( 2 t 4 − 4 t 2 ) dt = 5 2 t 5 − 3 4 t 3 + C = 5 2 ( x + 2 ) 5/2 − 3 4 ( x + 2 ) 3/2 + C
∫ x 1 + 2 x 2 dx
Sol. Let I = ∫ x 1 + 2 x 2 dx
Put 1 + 2 x 2 = t 2 ⇒ 4 xdx = 2 tdt ⇒ xdx = 2 t dt ∴ I = ∫ t . 2 t
dt = 2 1 ∫ t 2 dt
= 6 1 t 3 + C = 6 1 ( 1 + 2 x 2 ) 3/2 + C
∫ ( 4 x + 2 ) x 2 + x + 1 dx
Sol. Let I = ∫ ( 4 x + 2 ) x 2 + x + 1 dx
Put x 2 + x + 1 = t 2 ⇒ ( 2 x + 1 ) dx = 2 tdt
⇒ I = 2 ∫ t .2 tdt = 4 ∫ t 2 dt = 3 4 t 3 + C = 3 4 ( x 2 + x + 1 ) 3/2 + C
∫ x − x 1 dx
Sol. Let I = ∫ x − x 1 dx = ∫ x ( x − 1 ) 1 dx Put ( x − 1 ) = t ⇒ 2 x 1 dx = dt ⇒ x 1 dx = 2 dt
∴ I = ∫ t 1 .2 dt = 2 log ∣ t ∣ + C = 2 log ∣ x − 1∣ + C
∫ x + 4 x ⋅ dx , x > 0
Sol. Let I = ∫ x + 4 x ⋅ dx , x > 0
Put x + 4 = t 2 ⇒ x = t 2 − 4 ⇒ dx = 2 tdt
∴ I = ∫ t ( t 2 − 4 ) .2 tdt = 2 ∫ ( t 2 − 4 ) dt = 2 [ 3 t 3 − 4 t ] + C = 3 2 t [ t 2 − 12 ] + C = 3 2 x + 4 ( x − 8 ) + C
∫ ( x 3 − 1 ) 3 1 x 5 dx
Sol. Let I = ∫ ( x 3 − 1 ) 3 1 x 5 dx I = ∫ ( x 3 − 1 ) 1/3 ⋅ x 3 ⋅ x 2 d x
Put x 3 − 1 = t 3 ⇒ x 3 = t 3 + 1 ⇒ x 2 dx = t 2 dt
∴ I = ∫ t . ( t 3 + 1 ) . t 2 dt
= ∫ ( t 6 + t 3 ) d t = 7 t 7 + 4 t 4 + C = 7 1 ( x 3 − 1 ) 7/3 + 4 1 ( x 3 − 1 ) 4/3 + C
∫ ( 2 + 3 x 3 ) 3 x 2 dx
Sol. Let I = ∫ ( 2 + 3 x 3 ) 3 x 2 dx
Put 2 + 3 x 3 = t ⇒ 9 x 2 dx = dt ∴ I = ∫ t 3 1 . 9 dt = 9 1 ∫ t 3 1 dt = 9 1 [ − 2 t − 2 ] + C
= − 18 1 [ t 2 1 ] + C = − 18 ( 2 + 3 x 3 ) 2 1 + C
∫ x ( l o g x ) m 1 dx , x > 0 , m = 1
Sol. Let I = ∫ x ( l o g x ) m 1 dx Put log x = t ⇒ x 1 dx = dt
⇒
I = ∫ ( t ) m d t = ∫ t − m d t = ( 1 − m t 1 − m ) + C = ( 1 − m ) ( log x ) 1 − m + C
∫ 9 − 4 x 2 x dx
Sol. Let I = ∫ 9 − 4 x 2 x dx Put 9 − 4 x 2 = t ⇒ − 8 x d x = d t
∴
I = 8 − 1 ∫ t 1 dt = 8 − 1 log ∣ t ∣ + C = 8 − 1 log 9 − 4 x 2 + C
∫ e 2 x + 3 dx
Sol. Let I = ∫ e 2 x + 3 dx Put 2 x + 3 = t ⇒ dx = 2 dt
∴
I = 2 1 ∫ e t d t = 2 1 ( e t ) + C = 2 1 e ( 2 x + 3 ) + C
∫ e x 2 x dx Sol. Let I = ∫ e x 2 x dx
Put x 2 = t ⇒ 2 x d x = d t
∴
I = 2 1 ∫ e t 1 d t = 2 1 ∫ e − t d t = 2 1 ( − 1 e − t ) + C = − 2 1 e − x 2 + C = 2 e x 2 − 1 + C
∫ 1 + x 2 e t a n − 1 x dx
Sol. Let I = ∫ 1 + x 2 e t a n − 1 x dx Put tan − 1 x = t ⇒ 1 + x 2 1 dx = dt
∴ I = ∫ e t dt = e t + C = e tan − 1 x + C
∫ e 2 x + 1 e 2 x − 1 dx
Sol. Let I = ∫ e 2 x + 1 e 2 x − 1 dx
⇒ I = ∫ e x + e − x e x − e − x dx
Dividing numerator and denominator by ex Put e x + e − x = t ⇒ ( e x − e − x ) d x = d t
∴
I = ∫ t d t = log ∣ t ∣ + C = log e x + e − x + C
∫ e 2 x + e − 2 x e 2 x − e − 2 x dx
Sol. Let I = ∫ e 2 x + e − 2 x e 2 x − e − 2 x dx Put e 2 x + e − 2 x = t ⇒ ( 2 e 2 x − 2 e − 2 x ) d x = d t
⇒
( e 2 x − e − 2 x ) d x = 2 d t
∴
I = 2 1 ∫ t 1 d t
= 2 1 log ∣ t ∣ + C = 2 1 log e 2 x + e − 2 x + C
∫ tan 2 ( 2 x − 3 ) d x Sol.
Let I = ∫ tan 2 ( 2 x − 3 ) d x Put 2 x − 3 = t ⇒ d x = 2 d t
I = 2 1 ∫ tan 2 t ⋅ dt = 2 1 ∫ ( sec 2 t − 1 ) dt = 2 1 [ tan t − t ] + C 1 = 2 1 [ tan ( 2 x − 3 ) − ( 2 x − 3 )] + C 1 = 2 1 tan ( 2 x − 3 ) − x + 2 3 + C 1 = 2 1 tan ( 2 x − 3 ) − x + C ; where C = C 1 + 2 3
∫ sec 2 ( 7 − 4 x ) dx Sol. Let I = ∫ sec 2 ( 7 − 4 x ) dx
Put 7 − 4 x = t ⇒ − 4 dx = dt
∴ I = − 4 1 ∫ sec 2 tdt = 4 − 1 ( tan t ) + C = 4 − 1 tan ( 7 − 4 x ) + C
∫ 1 − x 2 s i n − 1 x dx Sol. Let I = ∫ 1 − x 2 s i n − 1 x dx
Put sin − 1 x = t ⇒ 1 − x 2 1 dx = dt
∴ I = ∫ tdt = 2 t 2 + C = 2 ( s i n − 1 x ) 2 + C
∫ 6 c o s x + 4 s i n x 2 c o s x − 3 s i n x d x Sol. Let I = ∫ 6 c o s x + 4 s i n x 2 c o s x − 3 s i n x dx
= 2 1 ∫ 3 c o s x + 2 s i n x 2 c o s x − 3 s i n x d x
Put 3 cos x + 2 sin x = t
⇒ ( − 3 sin x + 2 cos x ) d x = d t ⇒ ( 2 cos x − 3 sin x ) d x = d t
⇒ I = 2 1 ∫ t 1 dt = 2 1 log ∣ t ∣ + C = 2 1 log ∣2 sin x + 3 cos x ∣ + C
∫ c o s 2 x ( 1 − t a n x ) 2 1 dx Sol. Let I = ∫ c o s 2 x ( 1 − t a n x ) 2 1 dx
= ∫ ( 1 − t a n x ) 2 s e c 2 x d x
Put ( 1 − tan x ) = t ⇒ sec 2 x dx = − dt
∴ I = ∫ t 2 − d t = − ∫ t − 2 d t = t 1 + C = ( 1 − t a n x ) 1 + C
∫ x c o s x dx Sol. Let I = ∫ x c o s x dx
Put x = t ⇒ 2 x 1 dx = dt
∴ I = 2 ∫ cos t d t = 2 sin t + C = 2 sin x + C
∫ sin 2 x cos 2 x dx Sol. Let I = ∫ sin 2 x cos 2 x dx
Put sin 2 x = t 2
⇒ 2 cos 2 x d x = 2 t d t ⇒ cos 2 x d x = t d t
∴ I = ∫ t ⋅ t d t = ∫ t 2 d t = 3 t 3 + C = 3 1 ( sin 2 x ) 3/2 + C
∫ 1 + s i n x c o s x dx Sol. Let I = ∫ 1 + s i n x c o s x dx
Put 1 + sin x = t 2 ⇒ cos x dx = 2 t dt
∴ I = ∫ t 1 ⋅ 2 t d t = 2 ∫ 1 ⋅ d t = 2 t + C = 2 1 + sin x + C
∫ cot x log sin x dx Sol. Let I = ∫ cot x log sin x dx
Put log sin x = t
⇒ ∴ s i n x 1 ⋅ cos x d x = d t ⇒ cot x d x = d t I = ∫ t d t = 2 t 2 + C = 2 1 ( log sin x ) 2 + C
∫ 1 + c o s x s i n x dx Sol. Let I = ∫ 1 + c o s x s i n x dx
Put 1 + cos x = t ⇒ − sin x dx = dt
∴ I = ∫ − t d t = − log ∣ t ∣ + C = − log ∣1 + cos x ∣ + C
∫ ( 1 + c o s x ) 2 s i n x dx Sol. Let I = ∫ ( 1 + c o s x ) 2 s i n x dx
Put 1 + cos x = t ⇒ − sin x dx = dt
∴ I = ∫ − t 2 d t = − ∫ t − 2 d t = t 1 + C = 1 + cos x 1 + C
∫ 1 + c o t x 1 dx Sol. Let I = ∫ 1 + c o t x 1 dx
= ∫ 1 + ( s i n x c o s x ) 1 d x = ∫ sin x + cos x sin x d x = 2 1 ∫ sin x + cos x 2 sin x d x = 2 1 ∫ ( sin x + cos x ) ( sin x + cos x ) + ( sin x − cos x ) d x = 2 1 ∫ 1 d x − 2 1 ∫ sin x + cos x cos x − sin x d x = 2 x − 2 1 log ∣ sin x + cos x ∣ + C
[ ∵ ∫ f ( x ) f ′ ( x ) d x = log ∣ f ( x ) ∣ + C ]
∫ 1 − t a n x 1 dx Sol. Let I = ∫ 1 − t a n x 1 dx = ∫ 1 − ( c o s x s i n x ) 1 dx
= ∫ cos x − sin x cos x d x = 2 1 ∫ cos x − sin x 2 cos x d x = 2 1 ∫ ( cos x − sin x ) ( cos x − sin x ) + ( cos x + sin x ) d x = 2 1 ∫ 1 d x − 2 1 ∫ ( cos x − sin x ) ( − sin x − cos x ) d x = 2 x − 2 1 log ∣ cos x − sin x ∣ + C [ ∵ ∫ f ( x ) f ′ ( x ) d x = log ∣ f ( x ) ∣ + C ]
∫ s i n x c o s x t a n x dx Sol. Let I = ∫ s i n x c o s x t a n x dx = ∫ c o s x s i n x ⋅ c o s 2 x t a n x dx
= ∫ t a n x ⋅ c o s 2 x t a n x d x
⇒ I = ∫ t a n x s e c 2 xdx
Let tan x = t 2 ⇒ sec 2 x dx = 2 tdt
∴ I = ∫ t 1 ⋅ 2 t d t = 2 ∫ 1 d t = 2 t + C = 2 tan x + C
∫ x ( 1 + l o g x ) 2 dx Sol. Let I = ∫ x ( 1 + l o g x ) 2 dx
Put 1 + log x = t ⇒ x 1 dx = dt
∴ I = ∫ t 2 d t = 3 t 3 + C = 3 ( 1 + l o g x ) 3 + C
∫ x ( x + 1 ) ( x + l o g x ) 2 dx
Sol. Let I = ∫ x ( x + 1 ) ( x + l o g x ) 2 dx
Put ( x + log x ) = t
⇒ ( 1 + x 1 ) dx = dt ⇒ ( x x + 1 ) dx = dt
∴ I = ∫ t 2 dt = 3 t 3 + C
= 3 1 ( x + log x ) 3 + C ∫ 1 + x 8 x 3 s i n ( t a n − 1 x 4 ) dx
Sol. Let I = ∫ 1 + x 8 x 3 s i n ( t a n − 1 x 4 ) dx
Put tan − 1 x 4 = t ⇒ ( 1 + x 8 ) 4 x 3 dx = dt
⇒ ( 1 + x 8 ) x 3 dx = 4 dt
∴ I = 4 1 ∫ sin t . dt = 4 1 ( − cos t ) + C
= − 4 1 cos ( tan − 1 x 4 ) + C
Choose the correct answer in the following Exercises 38 and 39
∫ x 10 + 1 0 x 10 x 9 + 1 0 x l o g e 10 dx equals:
(A) 1 0 x − x 10 + C
(B) 1 0 x + x 10 + C
(C) ( 1 0 x − x 10 ) − 1 + C
(D) log ( 1 0 x + x 10 ) + C
Sol. (D)
Let I = ∫ x 10 + 1 0 x 10 x 9 + 1 0 x l o g e 10 dx
Put x 10 + 1 0 x = t
⇒ ( 10 x 9 + 1 0 x log e 10 ) dx = dt
∴ I = ∫ t dt = log ∣ t ∣ + C
= log 1 0 x + x 10 + C
Hence, the correct Answer is (D) ∫ s i n 2 x c o s 2 x dx equals ?
(A) tan x + cot x + C
(B) tan x − cot x + C
(C) tan x ⋅ cot x + C
(D) tan x − cot 2 x + C
Sol. (B)
Let I = ∫ s i n 2 x c o s 2 x dx = ∫ s i n 2 x c o s 2 x 1 dx
= ∫ s i n 2 x c o s 2 x s i n 2 x + c o s 2 x dx [ ∵ sin 2 x + cos 2 = 1 ]
= ∫ s i n 2 x c o s 2 x s i n 2 x dx + ∫ s i n 2 x c o s 2 x c o s 2 x dx
= ∫ sec 2 x dx + ∫ cosec 2 x dx
= tan x − cot x + C
Hence, the correct Answer is (B). EXERCISE 7.3 Find the integrals of the function in exercises 1 to 22
∫ sin 2 ( 2 x + 5 ) dx
Sol. Let I = ∫ sin 2 ( 2 x + 5 ) dx = 2 1 ∫ 2 sin 2 ( 2 x + 5 ) dx I = 2 1 ∫ { 1 − cos ( 4 x + 10 )} dx = 2 1 ∫ 1 dx − 2 1 ∫ cos ( 4 x + 10 ) dx = 2 1 x − 2 1 ( 4 sin ( 4 x + 10 ) ) + C = 2 1 x − 8 1 sin ( 4 x + 10 ) + C
∫ sin 3 x cos 4 x d x
Sol. Let I = ∫ sin 3 x cos 4 x dx = 2 1 ∫ 2 sin 3 x cos 4 x d x = 2 1 ∫ { sin 7 x + sin ( − x )} d x = 2 1 ∫ { sin 7 x − sin x } d x = 2 1 ∫ sin 7 x d x − 2 1 ∫ sin x d x = 2 1 ( 7 − cos 7 x ) − 2 1 ( − cos x ) + C = 14 − cos 7 x + 2 cos x + C
∫ cos 2 x cos 4 x cos 6 x d x Sol. Let
I = ∫ cos 2 x cos 4 x cos 6 x d x = 2 1 ∫ cos 2 x ( 2 cos 4 x cos 6 x ) d x = 2 1 ∫ cos 2 x [ cos ( 4 x + 6 x ) + cos ( 4 x − 6 x )] d x = 2 1 ∫ { cos 2 x cos 10 x + cos 2 x cos ( − 2 x )} d x [ ∵ cos ( − θ ) = cos θ ] = 2 1 ∫ { cos 2 x cos 10 x + cos 2 2 x } d x = 4 1 ∫ { 2 cos 2 x cos 10 x + 2 cos 2 2 x } d x = 4 1 ∫ ( cos 12 x + cos 8 x + 1 + cos 4 x ) d x = 4 1 [ 12 sin 12 x + 8 sin 8 x + x + 4 sin 4 x ] + C
∫ sin 3 ( 2 x + 1 ) dx Sol. Let I = ∫ sin 3 ( 2 x + 1 ) dx
= ∫ { 1 − cos 2 ( 2 x + 1 ) } sin ( 2 x + 1 ) dx
Put cos ( 2 x + 1 ) = t
⇒ ⇒ ∴ − 2 sin ( 2 x + 1 ) d x = d t sin ( 2 x + 1 ) d x = − 2 d t I = 2 − 1 ∫ ( 1 − t 2 ) d t = 2 − 1 { t − 3 t 3 } + C = 2 − 1 { cos ( 2 x + 1 ) − 3 c o s 3 ( 2 x + 1 ) } + C = 2 − c o s ( 2 x + 1 ) + 6 c o s 3 ( 2 x + 1 ) + C
∫ sin 3 x cos 3 x dx Sol. Let I = ∫ sin 3 x cos 3 x . dx
= ∫ cos 3 x ⋅ sin 2 x ⋅ sin x ⋅ d x = ∫ cos 3 x ( 1 − cos 2 x ) sin x ⋅ d x
Put cos x = t ⇒ − sin x . dx = dt
∴ I = − ∫ t 3 ( 1 − t 2 ) dt = − ∫ ( t 3 − t 5 ) dt = − { 4 t 4 − 6 t 6 } + C = − { 4 cos 4 x − 6 cos 6 x } + C = 6 cos 6 x − 4 cos 4 x + C
∫ sin x sin 2 x sin 3 x dx Sol. Let
I = ∫ sin x sin 2 x sin 3 x dx = 2 1 ∫ sin x { 2 sin 2 x sin 3 x } dx
∴ I = 2 1 ∫ [ sin x ⋅ { cos ( 2 x − 3 x ) − cos ( 2 x + 3 x )}] d x
= 4 1 ∫ ( 2 sin x cos x − 2 sin x cos 5 x ) d x
= 4 1 ∫ sin 2 x d x − 4 1 ∫ { sin 6 x − sin 4 x } d x
( ∵ ∫ sin axdx = − a c o s ax )
= 8 − c o s 2 x − 4 1 [ 6 − c o s 6 x + 4 c o s 4 x ] + C
= − 8 c o s 2 x − 8 1 [ − 3 c o s 6 x + 2 c o s 4 x ] + C
= 8 1 [ 3 c o s 6 x − 2 c o s 4 x − cos 2 x ] + C
∫ sin 4 x sin 8 x dx Sol. Let I = ∫ sin 4 x sin 8 x dx
= 2 1 ∫ 2 sin 4 x sin 8 x d x
∴ I = 2 1 ∫ { cos ( 4 x − 8 x ) − cos ( 4 x + 8 x )} d x
= 2 1 ∫ ( cos ( − 4 x ) − cos 12 x ) d x = 2 1 ∫ ( cos 4 x − cos 12 x ) d x = 2 1 [ 4 sin 4 x − 12 sin 12 x ] + C
∫ 1 + c o s x 1 − c o s x dx Sol. Let I = ∫ 1 + c o s x 1 − c o s x dx
= ∫ 2 cos 2 x /2 2 sin 2 x /2 d x = ∫ tan 2 2 x d x = ∫ ( sec 2 2 x − 1 ) d x = [ 2 1 tan 2 x − x ] + C = 2 tan 2 x − x + C
∫ 1 + c o s x c o s x dx Sol. Let I = ∫ 1 + c o s x c o s x dx
= ∫ 2 c o s 2 2 x c o s 2 2 x − s i n 2 2 x d x
∴ I = 2 1 ∫ ( 1 − tan 2 2 x ) dx
= 2 1 ∫ ( 1 − sec 2 2 x + 1 ) d x = 2 1 ∫ ( 2 − sec 2 2 x ) d x = 2 1 [ 2 x − 1/2 tan 2 x ] + C = x − tan 2 x + C
∫ sin 4 x dx Sol. Let I = ∫ sin 4 x dx = ∫ sin 2 x sin 2 x dx
= ∫ ( 2 1 − cos 2 x ) ( 2 1 − cos 2 x ) d x = ∫ 4 1 ( 1 − cos 2 x ) 2 d x = 4 1 ∫ [ 1 + cos 2 2 x − 2 cos 2 x ] d x = 4 1 ∫ [ 1 + ( 2 1 + cos 4 x ) − 2 cos 2 x ] d x = 4 1 ∫ [ 1 + 2 1 + 2 1 cos 4 x − 2 cos 2 x ] d x = 4 1 ∫ [ 2 3 + 2 1 cos 4 x − 2 cos 2 x ] d x = 8 3 x + 32 1 sin 4 x − 4 1 sin 2 x + C
∫ cos 4 2 x dx Sol. Let I = ∫ cos 4 2 x dx
= ∫ ( cos 2 2 x ) 2 d x = ∫ ( 2 1 + cos 4 x ) 2 d x = 4 1 ∫ [ 1 + cos 2 4 x + 2 cos 4 x ] d x = 4 1 ∫ [ 1 + ( 2 1 + cos 8 x ) + 2 cos 4 x ] d x = 4 1 ∫ [ 1 + 2 1 + 2 cos 8 x + 2 cos 4 x ] d x = 4 1 ∫ [ 2 3 + 2 cos 8 x + 2 cos 4 x ] d x = ∫ ( 8 3 + 8 cos 8 x + 2 cos 4 x ) d x = 8 3 x + 64 sin 8 x + 8 sin 4 x + C
∫ 1 + c o s x s i n 2 x dx Sol. Let I = ∫ 1 + c o s x s i n 2 x dx = ∫ ( 1 + c o s x ) ( 1 − c o s 2 x ) dx
= ∫ ( 1 − cos x ) d x = x − sin x + C
∫ c o s x − c o s α c o s 2 x − c o s 2 α d x
Sol. Let I = ∫ cos x − cos α cos 2 x − cos 2 α d x = ∫ ( cos x − cos α ) ( 2 cos 2 x − 1 ) − ( 2 cos 2 α − 1 ) d x = ∫ ( cos x − cos α ) 2 ( cos 2 x − cos 2 α ) d x = 2 ∫ ( cos x + cos α ) d x = 2 sin x + 2 x cos α + C = 2 [ sin x + x cos α ] + C
∫ 1 + s i n 2 x c o s x − s i n x dx
Sol. Let I = ∫ 1 + s i n 2 x c o s x − s i n x d x = ∫ ( s i n x + c o s x ) 2 c o s x − s i n x d x [ ∵ 1 + sin 2 x = ( sin x + cos x ) 2 ]
Put sin x + cos x = t ⇒ ( cos x − sin x ) dx = dt
∴ I = ∫ t 2 1 d t = − t 1 + C = − ( s i n x + c o s x ) 1 + C
∫ tan 3 2 x sec 2 x d x
Sol. Let I = ∫ tan 3 2 x sec 2 xdx = ∫ tan 2 2 x ⋅ sec 2 x tan 2 xdx = ∫ ( sec 2 2 x − 1 ) sec 2 x tan 2 xdx
Put sec 2 x = t
⇒ sec 2 x tan 2 xdx = 2 dt
∴
I = ∫ ( t 2 − 1 ) ⋅ 2 d t = 2 1 ∫ ( t 2 − 1 ) d t = 2 1 [ 3 t 3 − t ] + C = 6 1 ( sec 2 x ) 3 − 2 1 ( sec 2 x ) + C = 6 1 sec 3 2 x − 2 1 sec 2 x + C
∫ tan 4 x dx
Sol. Let I = ∫ tan 4 x d x = ∫ tan 2 x ⋅ tan 2 x d x = ∫ ( sec 2 x − 1 ) tan 2 x d x = ∫ ( tan 2 x sec 2 x − tan 2 x ) d x = ∫ tan 2 x sec 2 x d x − ∫ sec 2 x d x + ∫ 1 d x = 3 tan 3 x − tan x + x + C [ ∵ ∫ { f ( x ) } n ⋅ f ′ ( x ) d x = n + 1 { f ( x ) } n + 1 + C ]
∫ s i n 2 x c o s 2 x s i n 3 x + c o s 3 x dx
Sol. Let I = ∫ s i n 2 x c o s 2 x s i n 3 x + c o s 3 x dx = ∫ ( sin 2 x cos 2 x sin 3 x + sin 2 x cos 2 x cos 3 x ) d x = ∫ ( cos 2 x sin x + sin 2 x cos x ) d x = ∫ ( tan x sec x + cot x cosec x ) d x = sec x − cosec x + C
∫ c o s 2 x c o s 2 x + 2 s i n 2 x d x
Sol. Let I = ∫ c o s 2 x c o s 2 x + 2 s i n 2 x dx = ∫ cos 2 x ( 1 − 2 sin 2 x ) + 2 sin 2 x d x = ∫ cos 2 x 1 d x = ∫ sec 2 x d x = tan x + C
∫ s i n x c o s 3 x 1 dx Sol. Let I = ∫ s i n x c o s 3 x 1 dx
= ∫ sin x cos 3 x sin 2 x + cos 2 x d x = ∫ ( cos 3 x sin x + sin x cos x 1 ) d x = ∫ ( tan x sec 2 x + tan x sec 2 x ) d x
∴ I = ∫ tan x sec 2 x dx + ∫ t a n x s e c 2 x dx
Put tan x = t ⇒ sec 2 x dx = dt
⇒ I = ∫ tdt + ∫ t 1 dt
= 2 t 2 + log ∣ t ∣ + C = 2 1 tan 2 x + log ∣ tan x ∣ + C
∫ ( c o s x + s i n x ) 2 c o s 2 x d x Sol. Let I = ∫ ( c o s x + s i n x ) 2 c o s 2 x dx
= ∫ ( cos x + sin x ) 2 ( cos 2 x − sin 2 x ) d x = ∫ ( cos x + sin x ) 2 ( cos x + sin x ) ( cos x − sin x ) d x = ∫ ( cos x + sin x ) ( cos x − sin x ) d x = log ∣ sin x + cos x ∣ + C
[ ∵ ∫ f ( x ) f ′ ( x ) d x = log ∣ f ( x ) ∣ + C ]
∫ sin − 1 ( cos x ) dx Sol. Let I = ∫ sin − 1 ( cos x ) dx
= ∫ sin − 1 [ sin ( 2 π − x ) ] d x = ∫ ( 2 π − x ) d x = 2 π x − 2 x 2 + C
∫ c o s ( x − a ) c o s ( x − b ) 1 d x Sol. Let I = ∫ c o s ( x − a ) c o s ( x − b ) 1 dx
= sin ( a − b ) 1 ∫ [ cos ( x − a ) cos ( x − b ) sin ( a − b ) ] d x = sin ( a − b ) 1 ∫ [ cos ( x − a ) cos ( x − b ) sin [( x − b ) − ( x − a )] ] d x = sin ( a − b ) 1 ∫ [ cos ( x − a ) cos ( x − b ) [ sin ( x − b ) cos ( x − a ) − cos ( x − b ) sin ( x − a )] ] d x = sin ( a − b ) 1 ∫ [ tan ( x − b ) − tan ( x − a )] d x = sin ( a − b ) 1 [ ∵ ∫ tan x d x = − log ∣ cos x ∣ ) = sin ( a − b ) 1 [ cos ( x − b ) ∣ + log ∣ cos ( x − a ) ∣ ] cos ( x − a ) ∣ ] + C
Choose the correct answer in the following Exercises 23 and 24
23. ∫ s i n 2 x c o s 2 x s i n 2 x − c o s 2 x d x is equal to ?
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) − tan x + cot x + C
(D) tan x + sec x + C
Sol. (A)
Let I = ∫ s i n 2 x c o s 2 x s i n 2 x − c o s 2 x dx
= ∫ ( sin 2 x cos 2 x sin 2 x − sin 2 x cos 2 x cos 2 x ) d x = ∫ ( sec 2 x − cosec 2 x ) d x = tan x + cot x + C
Hence, the correct Answer is (A).
∫ c o s 2 ( e x x ) e x ( 1 + x ) d x equals ?
(A) − cot ( ex x ) + C
(B) tan ( xe x ) + C
(C) tan ( e x ) + C
(D) cot ( e x ) + C
Sol. (B) Let I = ∫ c o s 2 ( e x x ) e x ( 1 + x ) d x
Put e x . x = t ⇒ ( e x . x + e x .1 ) dx = dt
⇒ e x ( x + 1 ) dx = dt ∴ I = ∫ cos 2 t dt = ∫ sec 2 t dt = tan t + C = tan ( e x ⋅ x ) + C = tan ( x ⋅ e x ) + C
Hence, the correct Answer is (B).
EXERCISE 7.4 Integrate the functions in exercises 1 to 23
∫ x 6 + 1 3 x 2 d x
Sol. Let I = ∫ x 6 + 1 3 x 2 dx ∴ Put x 3 = t ⇒ 3 x 2 dx = dt I = ∫ ( x 3 ) 2 + 1 3 x 2 dx = ∫ t 2 + 1 dt = tan − 1 t + C = tan − 1 ( x 3 ) + C
∫ 1 + 4 x 2 1 dx
Sol. Let I = ∫ 1 + 4 x 2 1 dx
Put 2 x = t ⇒ 2 dx = dt ∴ I = ∫ 1 + ( 2 x ) 2 1 d x = 2 1 ∫ 1 + t 2 d t = 2 1 [ log t + t 2 + 1 ] + C = 2 1 log 2 x + 4 x 2 + 1 + C
∫ ( 2 − x ) 2 + 1 1 dx
Sol. Let I = ∫ ( 2 − x ) 2 + 1 1 dx
Put 2 − x = t ⇒ − dx = dt ∴ I = − ∫ t 2 + 1 1 dt = − log t + t 2 + 1 + C = − log ( 2 − x ) + ( 2 − x ) 2 + 1 + C = log ( 2 − x ) + x 2 − 4 x + 5 1 + C
∫ 9 − 25 x 2 1 d x
Sol. Let I = ∫ 9 − 25 x 2 1 dx ∴ Put 5 x = t ⇒ 5 dx = dt I = 5 1 ∫ 9 − t 2 1 dt = 5 1 ∫ 3 2 − t 2 1 dt = 5 1 sin − 1 ( 3 t ) + C = 5 1 sin − 1 ( 3 5 x ) + C
∫ 1 + 2 x 4 3 x d x
Sol. Let I = ∫ 1 + 2 x 4 3 x dx ∴ Put 2 x 2 = t ⇒ 2 2 xdx = dt I = 2 2 3 ∫ 1 + t 2 dt = 2 2 3 ( tan − 1 t ) + C = 2 2 3 tan − 1 ( 2 x 2 ) + C
∫ 1 − x 6 x 2 dx
Sol. Let I = ∫ 1 − x 6 x 2 dx ⇒ I = ∫ 1 − ( x 3 ) 2 x 2 dx Put x 3 = t ⇒ 3 x 2 dx = dt = 3 1 ∫ 1 − t 2 dt = 3 1 [ 2 1 log 1 − t 1 + t ] + C = 6 1 log 1 − x 3 1 + x 3 + C
∫ x 2 − 1 x − 1 dx
Sol. Let I = ∫ x 2 − 1 x − 1 dx ⇒ ∫ x 2 − 1 x d x − ∫ x 2 − 1 1 d x = ∫ x 2 − 1 x d x − log x + x 2 − 1
Put x 2 − 1 = t 2 ⇒ 2 xdx = 2 tdt
∴ I = ∫ t tdt − log x + x 2 − 1 = ∫ 1 dt − log x + x 2 − 1 = t − log x + x 2 − 1 = x 2 − 1 − log x + x 2 − 1 + C
∫ x 6 + a 6 x 2 dx
Sol. Let I = ∫ x 6 + a 6 x 2 dx ⇒ ∫ ( x 3 ) 2 + ( a 3 ) 2 x 2 d x Put x 3 = t ⇒ 3 x 2 d x = d t = 3 1 ∫ t 2 + ( a 3 ) 2 d t = 3 1 log t + t 2 + a 6 + C = 3 1 log x 3 + x 6 + a 6 + C
∫ t a n 2 x + 4 s e c 2 x dx
Sol. Let I = ∫ t a n 2 x + 4 s e c 2 x dx
Put tan x = t ⇒ sec 2 x dx = dt ∴ I = ∫ t 2 + 2 2 dt = log t + t 2 + 4 + C = log tan x + tan 2 x + 4 + C
∫ x 2 + 2 x + 2 1 dx
Sol. Let I = ∫ x 2 + 2 x + 2 1 dx = ∫ ( x + 1 ) 2 + 1 1 dx
Put x + 1 = t ⇒ dx = dt
∴ I = ∫ t 2 + 1 1 dt = log t + t 2 + 1 + C = log ( x + 1 ) + ( x + 1 ) 2 + 1 + C = log ( x + 1 ) + x 2 + 2 x + 2 + C
∫ ( 9 x 2 + 6 x + 5 ) 1 dx
Sol. Let I = ∫ ( 9 x 2 + 6 x + 5 ) 1 dx = ∫ ( 3 x + 1 ) 2 + 4 1 dx
Put ( 3 x + 1 ) = t ⇒ 3 dx = dt
∴ I = 3 1 ∫ t 2 + 2 2 1 dt = 3 1 [ 2 1 tan − 1 ( 2 t ) ] + C = 6 1 tan − 1 ( 2 3 x + 1 ) + C
∫ 7 − 6 x − x 2 1 dx Sol. Let I = ∫ 7 − 6 x − x 2 1 dx = ∫ 16 − ( x + 3 ) 2 1 dx
∴ Put x + 3 = t ⇒ dx = dt I = ∫ ( 4 ) 2 − ( t ) 2 1 dt = sin − 1 ( 4 t ) + C = sin − 1 ( 4 x + 3 ) + C
∫ ( x − 1 ) ( x − 2 ) 1 dx Sol. Let I = ∫ ( x − 1 ) ( x − 2 ) 1 dx
= ∫ x 2 − 3 x + 2 1 d x = ∫ ( x − 2 3 ) 2 − 4 1 1 d x
Put x − 2 3 = t ⇒ dx = dt
∴ I = ∫ t 2 − ( 2 1 ) 2 1 d t = log t + t 2 − ( 2 1 ) 2 + C
= log ( x − 2 3 ) + x 2 − 3 x + 2 + C
∫ 8 + 3 x − x 2 1 dx Sol. Let I = ∫ 8 + 3 x − x 2 1 dx
⇒ I = ∫ 4 41 − ( x − 2 3 ) 2 1 d x Put x − 2 3 = t ⇒ d x = d t
∴ I = ∫ ( 2 41 ) 2 − t 2 1 dt = sin − 1 ( 2 41 t ) + C
= sin − 1 ( 2 41 x − 3/2 ) + C = sin − 1 ( 41 2 x − 3 ) + C
∫ ( x − a ) ( x − b ) 1 dx Sol. Let I = ∫ ( x − a ) ( x − b ) 1 dx
= ∫ x 2 − ( a + b ) x + ab 1 d x = ∫ { x − ( 2 a + b ) } 2 − ( 2 a − b ) 2 1 d x
Put x − ( 2 a + b ) = t ⇒ dx = dt
⇒ ∫ { x − ( 2 a + b ) } 2 − ( 2 a − b ) 2 1 d x = ∫ t 2 − ( 2 a − b ) 2 1 d t = log t + t 2 − ( 2 a − b ) 2 + C = log ( x − 2 a + b ) + { x − ( 2 a + b ) } 2 − ( 2 a − b ) 2 + C = log { x − ( 2 a + b ) } + ( x − a ) ( x − b ) + C
∫ 2 x 2 + x − 3 4 x + 1 dx
Sol. Let I = ∫ 2 x 2 + x − 3 4 x + 1 dx
Put 2 x 2 + x − 3 = t 2 ⇒ ( 4 x + 1 ) dx = 2 t dt ∴ I = ∫ t 2 t d t = 2 ∫ 1 d t = 2 t + C = 2 2 x 2 + x − 3 + C
∫ x 2 − 1 x + 2 dx
Sol. Let I = ∫ x 2 − 1 x + 2 dx
Let x + 2 = A dx d ( x 2 − 1 ) + B ⇒ x + 2 = A ( 2 x ) + B
Equating the coefficients of x and constant term on both sides, we get 2 A = 1 ⇒ A = 2 1 , B = 2
From (1), we obtain, ( x + 2 ) = 2 1 ( 2 x ) + 2
∴ I ∴ I = ∫ x 2 − 1 2 1 ( 2 x ) + 2 dx = 2 1 ∫ x 2 − 1 2 x dx + ∫ x 2 − 1 2 dx = 2 1 ∫ x 2 − 1 2 x dx + 2 ∫ x 2 − 1 1 dx = 2 1 ∫ x 2 − 1 2 x dx + 2 log x + x 2 − 1 Put x 2 − 1 = t 2 ⇒ 2 xdx = 2 tdt = 2 1 ∫ t 2 t dt + 2 log x + x 2 − 1 = ∫ 1 dt + 2 log x + x 2 − 1 = t + 2 log x + x 2 − 1 + C = x 2 − 1 + 2 log x + x 2 − 1 + C
∫ 1 + 2 x + 3 x 2 5 x − 2 dx
Sol. Let I = ∫ 1 + 2 x + 3 x 2 5 x − 2 dx
Let 5 x − 2 = A dx d ( 1 + 2 x + 3 x 2 ) + B ⇒ 5 x − 2 = A ( 2 + 6 x ) + B
Equating the coefficients of x and constant term on both sides, we get
6 A = 5
⇒ A = 6 5 , 2 A + B = − 2 ⇒ B = − 2 − 2 A = − 3 11
⇒ ∴ 5 x − 2 = 6 5 ( 2 + 6 x ) − 3 11 I = ∫ 1 + 2 x + 3 x 2 6 5 ( 2 + 6 x ) − 3 11 dx = 6 5 ∫ 1 + 2 x + 3 x 2 2 + 6 x dx − 3 11 ∫ 1 + 2 x + 3 x 2 1 dx
Let I 1 = ∫ 1 + 2 x + 3 x 2 2 + 6 x dx and I 2 = ∫ 1 + 2 x + 3 x 2 1 dx
∴ I = 6 5 I 1 − 3 11 I 2
Now, I 1 = ∫ 1 + 2 x + 3 x 2 2 + 6 x dx = log 1 + 2 x + 3 x 2 + C 1
I 2 I 2 = ∫ 1 + 2 x + 3 x 2 1 d x = ∫ 3 [ x 2 + 3 2 x + 3 1 ] 1 d x = 3 1 ∫ [ ( x + 3 1 ) 2 + 9 2 ] 1 d x = 3 1 ∫ ( x + 3 1 ) 2 + ( 3 2 ) 1 d x = 3 1 ⋅ ( 2 /3 ) 1 tan − 1 { 2 /3 ( x + 1/3 ) } + C 2 = 2 1 tan − 1 ( 2 3 x + 1 ) + C 2
Using eq. (2) and (3) in eq. (1)
∴ I = 6 5 log 1 + 2 x + 3 x 2 + 6 5 C 1 − 3 11
[ 2 1 tan − 1 ( 2 3 x + 1 ) ] − 3 11 C 2
= 6 5 log 1 + 2 x + 3 x 2 − 3 2 11 tan − 1 ( 2 3 x + 1 ) + C
where C = 6 5 C 1 − 3 11 C 2
∫ ( x − 5 ) ( x − 4 ) 6 x + 7 dx
Sol. Let I = ∫ ( x − 5 ) ( x − 4 ) 6 x + 7 dx = ∫ x 2 − 9 x + 20 6 x + 7 d x
Let 6 x + 7 = A dx d ( x 2 − 9 x + 20 ) + B
⇒ 6 x + 7 = A ( 2 x − 9 ) + B
Equating the coefficients of x and constant term, we get
2 A = 6 ⇒ A = 3 − 9 A + B = 7 ⇒ B = 7 + 9 A = 34
∴ 6 x + 7 = 3 ( 2 x − 9 ) + 34
Let I 1 = ∫ x 2 − 9 x + 20 2 x − 9 dx
and I 2 = ∫ x 2 − 9 x + 20 1 dx
∴ I = 3 I 1 + 34 I 2
Now, I 1 = ∫ x 2 − 9 x + 20 2 x − 9 dx
Put x 2 − 9 x + 20 = t 2 ⇒ ( 2 x − 9 ) dx = 2 t dt
∴ I 1 = ∫ t 2 t dt = 2 ∫ 1 dt = 2 t + C 1 = 2 x 2 − 9 x + 20 + C 1
and
I 2 = ∫ x 2 − 9 x + 20 1 d x = ∫ ( x − 2 9 ) 2 − 4 1 1 d x
⇒ I 2 = log ( x − 2 9 ) + x 2 − 9 x + 20 + C 2
Using equations (2) and (3) in (1), we get
∴ I = 3 [ 2 x 2 − 9 x + 20 ] + 3 C 1 + 34 log
[ ( x − 2 9 ) + x 2 − 9 x + 20 ] + 34 C 2
I = 6 x 2 − 9 x + 20 + 34 log
[ ( x − 2 9 ) + x 2 − 9 x + 20 ] + C
where C = 3 C 1 + 34 C 2
∫ 4 x − x 2 x + 2 dx
Sol. Let I = ∫ 4 x − x 2 x + 2 dx
Let x + 2 = A dx d ( 4 x − x 2 ) + B
⇒ x + 2 = A ( 4 − 2 x ) + B Equating the coefficients of x and constant term on both sides, we get
− 2 A = 1 ⇒ A = 2 − 1 ⇒ 4 A + B = 2
⇒ B = 2 − 4 A = 4 ⇒ ( x + 2 ) = − 2 1 ( 4 − 2 x ) + 4
∴ I = ∫ 4 x − x 2 − 2 1 ( 4 − 2 x ) + 4 dx = − 2 1 ∫ 4 x − x 2 4 − 2 x dx + 4 ∫ 4 x − x 2 1 dx
Let I 1 = ∫ 4 x − x 2 4 − 2 x dx and I 2 = ∫ 4 x − x 2 1 dx
∴ I = − 2 1 I 1 + 4 I 2
Now, I 1 = ∫ 4 x − x 2 4 − 2 x dx
Put 4 x − x 2 = t 2 ⇒ ( 4 − 2 x ) dx = 2 tdt
∴ I 1 = ∫ t 2 t dt = 2 ∫ 1 dt = 2 t + C 1 = 2 4 x − x 2 + C 1
Again, I 2 = ∫ 4 x − x 2 1 dx
= ∫ 4 − ( x − 2 ) 2 1 d x = ∫ ( 2 ) 2 − ( x − 2 ) 2 1 d x = sin − 1 ( 2 x − 2 ) + C 2
Using equation's (2) and (3) in (1), we get
I = − 2 1 ( 2 4 x − x 2 ) − 2 1 C 1 + 4 sin − 1 ( 2 x − 2 ) + 4 C 2 = − 4 x − x 2 + 4 sin − 1 ( 2 x − 2 ) + C
where C = 4 C 2 − 2 1 C 1
∫ x 2 + 2 x + 3 x + 2 dx Sol. Let I = ∫ x 2 + 2 x + 3 x + 2 dx
= 2 1 ∫ x 2 + 2 x + 3 2 ( x + 2 ) d x = 2 1 ∫ x 2 + 2 x + 3 ( 2 x + 2 ) + 2 d x
= 2 1 ∫ x 2 + 2 x + 3 2 x + 2 dx + 2 1 ∫ x 2 + 2 x + 3 2 dx = 2 1 ∫ x 2 + 2 x + 3 2 x + 2 dx + ∫ x 2 + 2 x + 3 1 dx
Let I 1 = ∫ x 2 + 2 x + 3 2 x + 2 dx
and I 2 = ∫ x 2 + 2 x + 3 1 dx
∴ I = 2 1 I 1 + I 2
Now, I 1 = ∫ x 2 + 2 x + 3 2 x + 2 dx
Put x 2 + 2 x + 3 = t 2 ⇒ ( 2 x + 2 ) dx = 2 tdt
∴ I 1 = ∫ t 2 t d t = 2 ∫ 1 d t = 2 t + C 1
⇒ I 1 = 2 x 2 + 2 x + 3 + C 1
Again,
I 2 = ∫ x 2 + 2 x + 3 1 d x = ∫ ( x + 1 ) 2 + ( 2 ) 2 1 d x
I 2 = log ( x + 1 ) + x 2 + 2 x + 3 + C 2
Using equations (2) and (3) in (1), we get
I = 2 1 [ 2 x 2 + 2 x + 3 ] + 2 1 C 1 + log ( x + 1 ) + x 2 + 2 x + 3 + C 2 = x 2 + 2 x + 3 + log ( x + 1 ) + x 2 + 2 x + 3 + C
where C = 2 1 C 1 + C 2
∫ x 2 − 2 x − 5 x + 3 dx Sol. Let I = ∫ x 2 − 2 x − 5 x + 3 dx
Let x + 3 = A dx d ( x 2 − 2 x − 5 ) + B
⇒ x + 3 = A ( 2 x − 2 ) + B
Equating the coefficients of x and constant term on both sides, we obtain 2 A = 1 ⇒ A = 2 1
⇒ ∴ ∴ ∴ − 2 A + B = 3 ⇒ B = 3 + 2 A = 4 ( x + 3 ) = 2 1 ( 2 x − 2 ) + 4 I = ∫ x 2 − 2 x − 5 2 1 ( 2 x − 2 ) + 4 dx = 2 1 ∫ x 2 − 2 x − 5 ( 2 x − 2 ) dx + 4 ∫ x 2 − 2 x − 5 1 dx Let I 1 = ∫ x 2 − 2 x − 5 2 x − 2 dx and I 2 = ∫ x 2 − 2 x − 5 1 dx I = 2 1 I 1 + 4 I 2
Now, I 1 = ∫ x 2 − 2 x − 5 2 x − 2 dx
Put x 2 − 2 x − 5 = t
⇒ ( 2 x − 2 ) dx = dt = ∫ t dt = log ∣ t ∣ + C 1 I 1 = log x 2 − 2 x − 5 + C 1
Again, I 2 = ∫ x 2 − 2 x − 5 1 d x
= ∫ ( x 2 − 2 x + 1 ) − 6 1 d x = ∫ ( x − 1 ) 2 − ( 6 ) 2 1 d x = 2 6 1 log ( x − 1 + 6 x − 1 − 6 ) + C 2
Using equations (2) and (3) in (1), we get:
I = 2 1 log x 2 − 2 x − 5 + 2 1 C 1 + 2 6 4 log x − 1 + 6 x − 1 − 6 + 4 C 2 = 2 1 log x 2 − 2 x − 5 + 6 2 log x − 1 + 6 x − 1 − 6 + C
where C = 2 1 C 1 + 4 C 2
∫ x 2 + 4 x + 10 5 x + 3 dx Sol. Let I = ∫ x 2 + 4 x + 10 5 x + 3 dx
Let 5 x + 3 = A dx d ( x 2 + 4 x + 10 ) + B
⇒ 5 x + 3 = A ( 2 x + 4 ) + B
Equating the coefficients of x and constant term, we get
⇒ ⇒ ∴ 2 A = 5 ⇒ A = 2 5 , 4 A + B = 3 B = 3 − 4 ( 2 5 ) = − 7 5 x + 3 = 2 5 ( 2 x + 4 ) − 7 I = ∫ x 2 + 4 x + 10 2 5 ( 2 x + 4 ) − 7 dx = 2 5 ∫ x 2 + 4 x + 10 2 x + 4 dx − 7 ∫ x 2 + 4 x + 10 1 dx
Let, I 1 = ∫ x 2 + 4 x + 10 2 x + 4 dx
and I 2 = ∫ x 2 + 4 x + 10 1 dx
∴ I = 2 5 I 1 − 7 I 2
Now, I 1 = ∫ x 2 + 4 x + 10 2 x + 4 dx
Put x 2 + 4 x + 10 = t 2 ⇒ ( 2 x + 4 ) dx = 2 tdt
∴ I 1 = ∫ t 2 t d t = 2 ∫ 1 d t = 2 t + C 1 = 2 x 2 + 4 x + 10 + C 1
Again, I 2 = ∫ x 2 + 4 x + 10 1 dx
= ∫ ( x 2 + 4 x + 4 ) + 6 1 d x = ∫ ( x + 2 ) 2 + ( 6 ) 2 1 d x
Using equations (2) and (3) in (1), we get
I = 2 5 [ 2 x 2 + 4 x + 10 ] + 2 5 C 1 = 5 x 2 + 4 x + 10 ( x + 2 ) + x 2 + 4 x + 10 − 7 C 2 − 7 log ( x + 2 ) + x 2 + 4 x + 10 + C where C = 2 5 C 1 − 7 C 2
Choose the correct answer in the following Exercises 24 and 25
∫ x 2 + 2 x + 2 dx equals ?
(A) x tan − 1 ( x + 1 ) + C
(B) tan − 1 ( x + 1 ) + C
(C) ( x + 1 ) tan − 1 x + C
(D) tan − 1 x + C
Sol. (B) Let I = ∫ x 2 + 2 x + 2 dx = ∫ ( x 2 + 2 x + 1 ) + 1 dx = ∫ ( x + 1 ) 2 + ( 1 ) 2 1 dx = [ tan − 1 ( x + 1 ) ] + C
Hence, the correct Answer is (B).
∫ 9 x − 4 x 2 dx equals ?
(A) 9 1 sin − 1 ( 8 9 x − 8 ) + C
(B) 2 1 sin − 1 ( 9 8 x − 9 ) + C
(C) 3 1 sin − 1 ( 8 9 x − 8 ) + C
(D) 2 1 sin − 1 ( 9 9 x − 8 ) + C
Sol. (B) Let I = ∫ 9 x − 4 x 2 dx = ∫ − 4 ( x 2 − 4 9 x ) 1 d x
= ∫ − 4 ( x 2 − 4 9 x + 64 81 − 64 81 ) 1 d x = ∫ − 4 [ ( x − 8 9 ) 2 − ( 8 9 ) 2 ] 1 d x = 2 1 ∫ ( 8 9 ) 2 − ( x − 8 9 ) 2 1 d x = 2 1 [ sin − 1 ( 8 9 x − 8 9 ) ] + C = 2 1 sin − 1 ( 9 8 x − 9 ) + C
Hence, the correct Answer is (B).
EXERCISE 7.5 Integrate the rational functions in exercises 1 to 21.
∫ ( x + 1 ) ( x + 2 ) x dx
Sol. Let I = ∫ ( x + 1 ) ( x + 2 ) x dx Let ( x + 1 ) ( x + 2 ) x = ( x + 1 ) A + ( x + 2 ) B
⇒ x = A ( x + 2 ) + B ( x + 1 )
In eq. (1)
∴ Put x = − 1 ⇒ A = − 1 x = − 2 ⇒ − B = − 2 ⇒ B = 2 ( x + 1 ) ( x + 2 ) x = ( x + 1 ) − 1 + ( x + 2 ) 2
∴ I = ∫ ( x + 1 ) − 1 dx + ∫ ( x + 2 ) 2 dx = − log ∣ x + 1∣ + 2 log ∣ x + 2∣ + C = log ( x + 2 ) 2 − log ∣ x + 1∣ + C = log ( x + 1 ) ( x + 2 ) 2 + C
∫ x 2 − 9 1 dx
Sol. Let I = ∫ x 2 − 9 1 dx = ∫ ( x + 3 ) ( x − 3 ) 1 dx
Let ( x + 3 ) ( x − 3 ) 1 = ( x + 3 ) A + ( x − 3 ) B ⇒ 1 = A ( x − 3 ) + B ( x + 3 )
From eq. (1)
Put x = − 3 ⇒ − 6 A = 1
⇒ A = 6 − 1 and x = 3 ⇒ 6 B = 1 ⇒ B = 6 1
∴ ( x + 3 ) ( x − 3 ) 1 = 6 ( x + 3 ) − 1 + 6 ( x − 3 ) 1
∴ I = ∫ ( 6 ( x + 3 ) − 1 + 6 ( x − 3 ) 1 ) dx
= − 6 1 log ∣ x + 3∣ + 6 1 log ∣ x − 3∣ + C
= 6 1 log ( x + 3 ) ( x − 3 ) + C
∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) 3 x − 1 dx
Sol. Let I = ∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) 3 x − 1 dx
Let ( x − 1 ) ( x − 2 ) ( x − 3 ) 3 x − 1 = ( x − 1 ) A + ( x − 2 ) B + ( x − 3 ) C
3 x − 1 = A ( x − 2 ) ( x − 3 )
+ B ( x − 1 ) ( x − 3 ) + C ( x − 1 ) ( x − 2 )
Put x = 1 , 2 , and 3 in equation (1), we get
A = 1 , B = − 5 , and C = 4 respectively
Now, ∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) 3 x − 1 dx
= ∫ ( ( x − 1 ) 1 − ( x − 2 ) 5 + ( x − 3 ) 4 ) d x = log ∣ x − 1∣ − 5 log ∣ x − 2∣ + 4 log ∣ x − 3∣ + C
∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) x dx
Sol. Let I = ∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) x dx
Let ( x − 1 ) ( x − 2 ) ( x − 3 ) x = ( x − 1 ) A + ( x − 2 ) B + ( x − 3 ) C
x = A ( x − 2 ) ( x − 3 )
+ B ( x − 1 ) ( x − 3 ) + C ( x − 1 ) ( x − 2 )
Put x = 1 , 2 and 3 in equation (1), we get
A = 2 1 , B = − 2 , and C = 2 3 respectively
∴ ( x − 1 ) ( x − 2 ) ( x − 3 ) x
= 2 ( x − 1 ) 1 − ( x − 2 ) 2 + 2 ( x − 3 ) 3
∴ I = ∫ { 2 ( x − 1 ) 1 − ( x − 2 ) 2 + 2 ( x − 3 ) 3 } dx
= 2 1 log ∣ x − 1∣ − 2 log ∣ x − 2∣ + 2 3 log ∣ x − 3∣ + C
∫ x 2 + 3 x + 2 2 x d x
Sol. Let I = ∫ x 2 + 3 x + 2 2 x dx , x 2 + 3 x + 2 2 x = ( x + 1 ) ( x + 2 ) 2 x = ( x + 1 ) A + ( x + 2 ) B
⇒ 2 x = A ( x + 2 ) + B ( x + 1 )
Put x = − 1 and -2 equation (1),
we get A = − 2 and B = 4 respectively
⇒ ( x + 1 ) ( x + 2 ) 2 x = ( x + 1 ) − 2 + ( x + 2 ) 4
⇒ I = ∫ { ( x + 2 ) 4 − ( x + 1 ) 2 } dx
= 4 log ∣ x + 2∣ − 2 log ∣ x + 1∣ + C
∫ x ( 1 − 2 x ) 1 − x 2 dx
Sol. Let I = ∫ x ( 1 − 2 x ) 1 − x 2 dx
Given integrand is improper rational function.
So, on dividing we get x ( 1 − 2 x ) 1 − x 2 = 2 1 + 2 1 ( x ( 1 − 2 x ) 2 − x )
Let x ( 1 − 2 x ) 2 − x = x A + ( 1 − 2 x ) B
⇒ 2 − x = A ( 1 − 2 x ) + B x
Put x = 0 and 2 1 in equation (1), we get A = 2 and B = 3 respectively
∴ x ( 1 − 2 x ) 2 − x = x 2 + 1 − 2 x 3
∴ I = ∫ { 2 1 + 2 1 ( x 2 + 1 − 2 x 3 ) } dx
= 2 x + log ∣ x ∣ + 2 ( − 2 ) 3 log ∣1 − 2 x ∣ + C
= 2 x + log ∣ x ∣ − 4 3 log ∣1 − 2 x ∣ + C
∫ ( x 2 + 1 ) ( x − 1 ) x dx
Sol. Let I = ∫ ( x 2 + 1 ) ( x − 1 ) x dx
Put ( x 2 + 1 ) ( x − 1 ) x = ( x 2 + 1 ) Ax + B + ( x − 1 ) C , x = ( Ax + B ) ( x − 1 ) + C ( x 2 + 1 )
In eq. (1), Put x = 1 ⇒ C = 2 1
Equating the coefficients of x 2 and constant term, we get A + C = 0
⇒ A = − C = − 2 1 , − B + C = 0 ⇒ B = C = 2 1 ∴ ( x 2 + 1 ) ( x − 1 ) x = x 2 + 1 ( − 2 1 x + 2 1 ) + ( x − 1 ) 2 1
∴ I = − 2 1 ∫ x 2 + 1 x dx + 2 1 ∫ x 2 + 1 1 dx + 2 1 ∫ x − 1 1
= − 4 1 ∫ x 2 + 1 2 x dx + 2 1 tan − 1 x + 2 1 log ∣ x − 1∣ + C = − 4 1 log x 2 + 1 + 2 1 tan − 1 x + 2 1 log ∣ x − 1∣ + C = 2 1 log ∣ x − 1∣ − 4 1 log x 2 + 1 + 2 1 tan − 1 x + C
∫ ( x − 1 ) 2 ( x + 2 ) x dx
Sol. Let I = ∫ ( x − 1 ) 2 ( x + 2 ) x dx
Let ( x − 1 ) 2 ( x + 2 ) x = ( x − 1 ) A + ( x − 1 ) 2 B + ( x + 2 ) C x = A ( x − 1 ) ( x + 2 ) + B ( x + 2 ) + C ( x − 1 ) 2
In eq. (1)
Put x = 1 ⇒ B = 3 1 , x = − 2 ⇒ C = − 9 2
On equating the coefficients of x 2 , A + C = 0
⇒ A = − C = 9 2 ∴ ( x − 1 ) 2 ( x + 2 ) x = 9 ( x − 1 ) 2 + 3 ( x − 1 ) 2 1 − 9 ( x + 2 ) 2 ∴ I = 9 2 ∫ ( x − 1 ) 1 d x + 3 1 ∫ ( x − 1 ) 2 1 d x − 9 2 ∫ ( x + 2 ) 1 d x = 9 2 log ∣ x − 1∣ + 3 1 ( x − 1 − 1 ) − 9 2 log ∣ x + 2∣ + C = 9 2 log x + 2 x − 1 − 3 ( x − 1 ) 1 + C
∫ x 3 − x 2 − x + 1 3 x + 5 dx
Sol. Let I = ∫ x 3 − x 2 − x + 1 3 x + 5 dx
Let ( x − 1 ) 2 ( x + 1 ) 3 x + 5 = ( x − 1 ) A + ( x − 1 ) 2 B + ( x + 1 ) C 3 x + 5 = A ( x − 1 ) ( x + 1 ) + B ( x + 1 ) + C ( x − 1 ) 2
Put x = 1 and x = − 1 in equation (1), we get
B = 4 and C = 2 1
Equating the coefficients of x 2 we get
A + C = 0 ⇒ A = − C = − 2 1
∴ ( x − 1 ) 2 ( x + 1 ) 3 x + 5 = 2 ( x − 1 ) − 1 + ( x − 1 ) 2 4 + 2 ( x + 1 ) 1
∴ I = − 2 1 ∫ x − 1 1 dx + 4 ∫ ( x − 1 ) 2 1 dx + 2 1 ∫ ( x + 1 ) 1 dx = − 2 1 log ∣ x − 1∣ + 4 ( x − 1 − 1 ) + 2 1 log ∣ x + 1∣ + C = 2 1 log x − 1 x + 1 − ( x − 1 ) 4 + C
∫ ( x 2 − 1 ) ( 2 x + 3 ) 2 x − 3 dx Sol. Let I = ∫ ( x 2 − 1 ) ( 2 x + 3 ) 2 x − 3 dx
= ∫ ( x + 1 ) ( x − 1 ) ( 2 x + 3 ) 2 x − 3 ⋅ dx
Let ( x + 1 ) ( x − 1 ) ( 2 x + 3 ) 2 x − 3 = ( x + 1 ) A + ( x − 1 ) B + ( 2 x + 3 ) C
⇒ ( 2 x − 3 ) = A ( x − 1 ) ( 2 x + 3 ) + B ( x + 1 ) ( 2 x + 3 ) + C ( x + 1 ) ( x − 1 )
Put x = − 1 , 1 and − 2 3 in eq. (1), we get
∴ ∴ = = A = 2 5 , B = − 10 1 , C = − 5 24 respectively ( x + 1 ) ( x − 1 ) ( 2 x + 3 ) 2 x − 3 = 2 ( x + 1 ) 5 − 10 ( x − 1 ) 1 − 5 ( 2 x + 3 ) 24 I = 2 5 ∫ ( x + 1 ) 1 d x − 10 1 ∫ x − 1 1 d x − 5 24 ∫ ( 2 x + 3 ) 1 d x 2 5 log ∣ x + 1∣ − 10 1 log ∣ x − 1∣ − 5 × 2 24 log ∣2 x + 3∣ + C 2 5 log ∣ x + 1∣ − 10 1 log ∣ x − 1∣ − 5 12 log ∣2 x + 3∣ + C
∫ ( x + 1 ) ( x 2 − 4 ) 5 x dx Sol. Let I = ∫ ( x + 1 ) ( x 2 − 4 ) 5 x dx
= ∫ ( x + 1 ) ( x + 2 ) ( x − 2 ) 5 x d x
Let ( x + 1 ) ( x + 2 ) ( x − 2 ) 5 x
= ( x + 1 ) A + ( x + 2 ) B + ( x − 2 ) C ( x + 1 ) ( x + 2 ) ( x − 2 ) 5 x = ( x + 1 ) A + ( x + 2 ) B + ( x − 2 ) C 5 x = A ( x + 2 ) ( x − 2 ) + B ( x + 1 ) ( x − 2 ) + C ( x + 1 ) ( x + 2 )
Put x = − 1 , − 2 , and 2 in equation (1), we get
∴ ∴ A = 3 5 , B = − 2 5 and C = 6 5 respectively ( x + 1 ) ( x + 2 ) ( x − 2 ) 5 x = 3 ( x + 1 ) 5 − 2 ( x + 2 ) 5 + 6 ( x − 2 ) 5 I = 3 5 ∫ ( x + 1 ) 1 dx − 2 5 ∫ ( x + 2 ) 1 dx + 6 5 ∫ ( x − 2 ) 1 dx = 3 5 log ∣ x + 1∣ − 2 5 log ∣ x + 2∣ + 6 5 log ∣ x − 2∣ + C
∫ x 2 − 1 x 3 + x + 1 dx Sol. Let I = ∫ x 2 − 1 x 3 + x + 1 dx
Given integrand is improper rational function.
So, on dividing ( x 3 + x + 1 ) by x 2 − 1 , we get
x 2 − 1 x 3 + x + 1 = x + x 2 − 1 2 x + 1
Let x 2 − 1 2 x + 1 = ( x + 1 ) A + ( x − 1 ) B
⇒ 2 x + 1 = A ( x − 1 ) + B ( x + 1 )
Put x = − 1 and 1 in equation (1), we get
A = 2 1 and B = 2 3
∴ x 2 − 1 x 3 + x + 1 = x + 2 ( x + 1 ) 1 + 2 ( x − 1 ) 3
∴ I = ∫ x d x + 2 1 ∫ ( x + 1 ) 1 d x + 2 3 ∫ ( x − 1 ) 1 d x = 2 x 2 + 2 1 log ∣ x + 1∣ + 2 3 log ∣ x − 1∣ + C
∫ ( 1 − x ) ( 1 + x 2 ) 2 dx Sol. Let I = ∫ ( 1 − x ) ( 1 + x 2 ) 2 dx
= ∫ ( 1 − x ) ( 1 + x 2 ) 2 d x
Let ( 1 − x ) ( 1 + x 2 ) 2 = ( 1 − x ) A + ( 1 + x 2 ) B x + C
2 = A ( 1 + x 2 ) + ( B x + C ) ( 1 − x )
In eq. (1), Put x = 1 ⇒ A = 1
Now, on equating the coefficient of x 2 and constant term, we get
∴ A − B = 0 ⇒ B = A = 1 A + C = 2 ⇒ C = 2 − A = 1 ( 1 − x ) ( 1 + x 2 ) 2 = 1 − x 1 + 1 + x 2 x + 1
∴ I = ∫ 1 − x 1 d x + ∫ 1 + x 2 x d x + ∫ 1 + x 2 1 d x = − ∫ x − 1 1 d x + 2 1 ∫ 1 + x 2 2 x d x + ∫ 1 + x 2 1 d x = − log ∣ x − 1∣ + 2 1 log 1 + x 2 + tan − 1 x + C
∫ ( x + 2 ) 2 3 x − 1 dx Sol. Let I = ∫ ( x + 2 ) 2 3 x − 1 dx = ∫ ( x + 2 ) 2 3 x − 1 dx
Let ( x + 2 ) 2 3 x − 1 = ( x + 2 ) A + ( x + 2 ) 2 B
⇒ 3 x − 1 = A ( x + 2 ) + B
In eq. (1), Put x = − 2 ⇒ B = − 7
On equating the coefficients of x, we get A = 3
∴ ⇒ ( x + 2 ) 2 3 x − 1 = ( x + 2 ) 3 − ( x + 2 ) 2 7 I = 3 ∫ ( x + 2 ) 1 dx − 7 ∫ ( x + 2 ) 2 1 dx = 3 log ∣ x + 2∣ − 7 ( ( x + 2 ) − 1 ) + C = 3 log ∣ x + 2∣ + ( x + 2 ) 7 + C
∫ x 4 − 1 1 dx Sol. Let I = ∫ x 4 − 1 1 dx
⇒ ( x 4 − 1 ) 1 = ( x 2 − 1 ) ( x 2 + 1 ) 1 = ( x + 1 ) ( x − 1 ) ( 1 + x 2 ) 1
Let ( x + 1 ) ( x − 1 ) ( x 2 + 1 ) 1
= ( x + 1 ) A + ( x − 1 ) B + ( x 2 + 1 ) Cx + D
1 = A ( x − 1 ) ( x 2 + 1 ) + B ( x + 1 ) ( x 2 + 1 ) + ( Cx + D ) ( x 2 − 1 )
In eq. (1)
Put x = − 1 and 1 we get A = − 4 1 and B = 4 1
On equating the coefficients of x 3 and constant term, we get
∴ ∴ A + B + C = 0 ⇒ C = − ( A + B ) = 0 − A + B − D = 1 ⇒ D = − A + B − 1 = − 2 1 x 4 − 1 1 = 4 ( x + 1 ) − 1 + 4 ( x − 1 ) 1 − 2 ( x 2 + 1 ) 1 I = − 4 1 ∫ x + 1 1 dx + 4 1 ∫ x − 1 1 dx − 2 1 ∫ x 2 + 1 1 dx = − 4 1 log ∣ x + 1∣ + 4 1 log ∣ x − 1∣ − 2 1 tan − 1 x + C = 4 1 log x + 1 x − 1 − 2 1 tan − 1 x + C
∫ x ( x n + 1 ) 1 dx
Sol. Let I = ∫ x ( x n + 1 ) 1 dx x ( x n + 1 ) 1 = x n ( x n + 1 ) x n − 1
( ∵ Multiplying numerator and denominator by x n − 1 )
Put x n = t ⇒ nx n − 1 dx = dt
∴ I = ∫ x n ( x n + 1 ) x n − 1 dx = n 1 ∫ t ( t + 1 ) 1 dt
⇒ n 1 ∫ { t 1 − ( t + 1 ) 1 } dx = n 1 [ log ∣ t ∣ − log ∣ t + 1∣ ] + C
= n 1 [ log ∣ x n ∣ − log ∣ x n + 1 ∣ ] + C
= n 1 log x n + 1 x n + C
Alternate :
I = ∫ x ( x n + 1 ) 1 d x = ∫ x n + 1 ( 1 + x − n ) 1 d x
Put 1 + x − n = t ⇒ − nx − n − 1 dx = dt
⇒ x n + 1 1 d x = − n d t
∴ I = − n 1 ∫ t 1 dt = − n 1 log ∣ t ∣ + C
= − n 1 log ∣ 1 + x − n ∣ + C = n 1 log x n + 1 x n + C
∫ ( 1 − s i n x ) ( 2 − s i n x ) c o s x dx
Sol. Let I = ∫ ( 1 − s i n x ) ( 2 − s i n x ) c o s x dx
Put sin x = t ⇒ cos xdx = dt ∴ I = ∫ ( 1 − t ) ( 2 − t ) d t Let ( 1 − t ) ( 2 − t ) 1 = ( 1 − t ) A + ( 2 − t ) B 1 = A ( 2 − t ) + B ( 1 − t )
Put t = 1 and t = 2 in equation (1), we get A = 1 and B = − 1 respectively
∴ ( 1 − t ) ( 2 − t ) 1 = ( 1 − t ) 1 − ( 2 − t ) 1
⇒ I = ∫ { 1 − t 1 − ( 2 − t ) 1 } dx
= − log ∣1 − t ∣ + log ∣2 − t ∣ + C = log 1 − t 2 − t + C = log 1 − sin x 2 − sin x + C
∫ ( x 2 + 3 ) ( x 2 + 4 ) ( x 2 + 1 ) ( x 2 + 2 ) d x
Sol. Let I = ∫ ( x 2 + 3 ) ( x 2 + 4 ) ( x 2 + 1 ) ( x 2 + 2 ) dx , x 2 = y ,
then ( y + 3 ) ( y + 4 ) ( y + 1 ) ( y + 2 ) = 1 + ( y + 3 ) ( y + 4 ) ( − 4 y − 10 ) = 1 + y + 3 A + y + 4 B
⇒ ( y + 1 ) ( y + 2 )
= ( y + 3 ) ( y + 4 ) + A ( y + 4 ) + B ( y + 3 )
In eq. (1)
Put y = − 3 and y = − 4 , we get A = 2 and
B = − 6 respectively
∴ ( y + 3 ) ( y + 4 ) ( y + 1 ) ( y + 2 ) = 1 + ( y + 3 ) 2 − ( y + 4 ) 6
⇒ ( x 2 + 3 ) ( x 2 + 4 ) ( x 2 + 1 ) ( x 2 + 2 ) = 1 + x 2 + 3 2 − x 2 + 4 6 [ ∵ x 2 = y ]
∴ I = ∫ 1 dx + 2 ∫ x 2 + 3 1 dx − 6 ∫ x 2 + 4 1 dx
= x + 2 ⋅ 3 1 tan − 1 3 x − 6 ⋅ 2 1 tan − 1 2 x + C
= x + 3 2 tan − 1 3 x − 3 tan − 1 2 x + C
∫ ( x 2 + 1 ) ( x 2 + 3 ) 2 x dx
Sol. Let I = ∫ ( x 2 + 1 ) ( x 2 + 3 ) 2 x dx
Put x 2 = t ⇒ 2 x dx = dt
∴ I = ∫ ( t + 1 ) ( t + 3 ) dt = ∫ ( t + 1 ) ( t + 3 ) 1 dx
Let t + 1 A + t + 3 B
1 = ( t + 3 ) A + ( t + 1 ) B ,
1 = ( A + B ) t + ( 3 t + B )
⇒ A = 2 1 and B = − 2 1
∴ I = ∫ 2 1 [ ( t + 1 ) 1 − ( t + 3 ) 1 ] dt
= 2 1 ∫ t + 1 1 dt − 2 1 ∫ t + 3 1 dt
= 2 1 log ∣ t + 1∣ − 2 1 log ∣ t + 3∣ + C
= 2 1 log t + 3 t + 1 + C = 2 1 log x 2 + 3 x 2 + 1 + C
∫ x ( x 4 − 1 ) 1 dx
Sol. Let I = ∫ x ( x 4 − 1 ) 1 dx = ∫ x 4 ( x 4 − 1 ) x 3 dx
(Multiply Nr. and Dr. by x 3 )
Put x 4 = t ⇒ 4 x 3 dx = dt
∴ ∫ x ( x 4 − 1 ) 1 dx = 4 1 ∫ t ( t − 1 ) dt = 4 1 ∫ [ t − 1 1 − t 1 ] d t = 4 1 ∫ t − 1 1 d t − 4 1 ∫ t 1 d t = 4 1 log ∣ t − 1∣ − 4 1 log ∣ t ∣ + C = 4 1 log t t − 1 + C = 4 1 log x 4 x 4 − 1 + C
∫ ( e x − 1 ) 1 dx
Sol. Let I = ∫ ( e x − 1 ) 1 dx
Let e x = t ⇒ e x dx = dt ⇒ dx = t dt
∴ I = ∫ t − 1 1 × t d t = ∫ t ( t − 1 ) 1 d t
= ∫ [ t − 1 1 − t 1 ] dt = log ∣ t − 1∣ − log ∣ t ∣ + C
= log t t − 1 + C = log e x e x − 1 + C Choose the correct answer in the following Exercises 22 and 23
∫ ( x − 1 ) ( x − 2 ) xdx equals ?
(A) log x − 2 ( x − 1 ) 2 + C
(B) log x − 1 ( x − 2 ) 2 + C
(C) log ( x − 2 ( x − 1 ) ) 2 + C
(D) log ∣ ( x − 1 ) ( x − 2 ) ∣ + C
Sol. (B) Let I = ∫ ( x − 1 ) ( x − 2 ) x dx
Let ( x − 1 ) ( x − 2 ) x = ( x − 1 ) A + ( x − 2 ) B x = A ( x − 2 ) + B ( x − 1 )
Put x = 1 and 2 in (1), we get
A = − 1 and B = 2 respectively,
∴ ( x − 1 ) ( x − 2 ) x = − ( x − 1 ) 1 + ( x − 2 ) 2
⇒ I = ∫ { ( x − 1 ) − 1 + ( x − 2 ) 2 } dx
= − log ∣ x − 1∣ + 2 log ∣ x − 2∣ + C
= log ( x − 1 ) ( x − 2 ) 2 + C
Hence, the correct Answer is (B).
∫ x ( x 2 + 1 ) dx equals ?
(A) log ∣ x ∣ − 2 1 log ( x 2 + 1 ) + C
(B) log ∣ x ∣ + 2 1 log ( x 2 + 1 ) + C
(C) − log ∣ x ∣ + 2 1 log ( x 2 + 1 ) + C
(D) 2 1 log ∣ x ∣ + log ( x 2 + 1 ) + C
Sol. (A) Let I = ∫ x ( x 2 + 1 ) dx dx
Let x ( x 2 + 1 ) 1 = x A + x 2 + 1 Bx + C I = A ( x 2 + 1 ) + ( B x + C ) x
In eq. (1), Put x = 0 ⇒ A = 1
On equating the coefficients of x 2 , x , we get A + B = 0
⇒ B = − A = − 1 , A = 1 and C = 0
∴ x ( x 2 + 1 ) 1 = x 1 + ( x 2 + 1 − x )
Hence, the correct Answer is (A).
EXERCISE 7.6
Integrate the functions in Exercises 1 to 22.
∫ x sin x d x
Sol. Let I = ∫ x sin x dx
Taking x as first function and sin x as second function and integrating by parts, we obtain I = x ∫ sin x d x − ∫ { ( d x d ( x ) ) ∫ sin x d x } d x = x ( − cos x ) + ∫ 1 ⋅ ( cos x ) d x = − x cos x + sin x + C
Let ∫ x sin 3 x dx
Sol. I = ∫ x sin 3 x dx
Taking x as first function and sin 3 x as second function and integrating by parts, we obtain I = x ∫ sin 3 x d x − ∫ { ( d x d ( x ) ) ∫ sin 3 x d x } d x = x ( 3 − cos 3 x ) − ∫ 1 ⋅ ( 3 − cos 3 x ) d x = 3 − x cos 3 x + 3 1 ∫ cos 3 x d x = 3 − x cos 3 x + 9 1 sin 3 x + C
∫ x 2 e x d x
Sol. Let I = ∫ x 2 e x dx
Taking x 2 as first function and e x as second function and integrating by parts, we obtain I = x 2 ∫ e x d x − ∫ { ( d x d ( x 2 ) ) ∫ e x d x } d x = x 2 e x − ∫ 2 x ⋅ e x d x = x 2 e x − 2 ∫ x ⋅ e x d x
Again integrating by parts, we obtain
I = x 2 e x − 2 [ x ∫ e x d x − ∫ { ( d x d ( x ) ) ⋅ ∫ e x d x } d x ] = x 2 e x − 2 [ x e x − ∫ e x d x ] = x 2 e x − 2 [ x e x − e x ] = x 2 e x − 2 x e x + 2 e x + C = e x ( x 2 − 2 x + 2 ) + C
∫ x log x d x
Sol. Let I = ∫ x log x dx
Taking log x as first function and x as second function and integrating by parts, we obtain I = log x ∫ x d x − ∫ { ( d x d ( log x ) ) ∫ x d x } d x = log x ⋅ 2 x 2 − ∫ x 1 ⋅ 2 x 2 d x = 2 x 2 log x − ∫ 2 x d x = 2 x 2 log x − 4 x 2 + C
∫ x log 2 x dx
Sol. Let I = ∫ x log 2 x dx
Taking log 2 x as first function and x as second function and integrating by parts, we obtain I = log 2 x ∫ x d x − ∫ { ( d x d ( log 2 x ) ) ∫ x d x } d x = log 2 x ⋅ 2 x 2 − ∫ ( 2 x 2 ⋅ 2 x 2 ) d x = 2 x 2 log 2 x − ∫ 2 x d x = 2 x 2 log 2 x − 4 x 2 + C
∫ x 2 log x dx
Sol. Let I = ∫ x 2 log x dx
Taking log x as first function and x 2 as second function and integrating by parts, we obtain I = log x ∫ x 2 d x − ∫ { ( d x d ( log x ) ) ∫ x 2 d x } d x = log x ( 3 x 3 ) − ∫ x 1 ⋅ 3 x 3 d x = 3 x 3 log x − ∫ 3 x 2 d x = 3 x 3 log x − 9 x 3 + C
∫ x sin − 1 x d x
Sol. Let I = ∫ x sin − 1 x dx
Put x = sin t , dx = cos tdt = ∫ sin t sin − 1 ( sin t ) cos t d t = ∫ t sin t cos t d t = 2 1 ∫ t ( 2 sin t cos t ) d t = 2 1 ∫ t sin 2 t d t
Taking t as first function and sin 2 t as second function and integrating by parts, we obtain
= 2 1 [ t ∫ sin 2 t d t − ∫ ( d t d ( t ) ∫ sin 2 t d t ) d t ] = 2 1 [ − t 2 cos 2 t + ∫ 2 cos 2 t d t ]
= 2 1 [ − 2 t cos 2 t + 4 1 sin 2 t ] + C = 4 − t cos 2 t + 8 1 sin 2 t + C = 4 − t [ 1 − 2 sin 2 t ] + 8 1 2 sin t cos t + C = 4 − t ( 1 − 2 sin 2 t ) + 4 1 sin t 1 − sin 2 t + C = 4 − sin − 1 x ( 1 − 2 x 2 ) + 4 x 1 − x 2 + C = 4 1 ( 2 x 2 − 1 ) sin − 1 x + 4 x 1 − x 2 + C
∫ x tan − 1 x dx
Sol. I = ∫ x tan − 1 x dx
Taking tan − 1 x as first function and x as second function and integrating by parts, we obtain I = tan − 1 x ∫ x d x − ∫ { ( d x d ( tan − 1 x ) ) ∫ x d x } d x = tan − 1 x ( 2 x 2 ) − ∫ 1 + x 2 1 ⋅ 2 x 2 d x = 2 x 2 tan − 1 x − 2 1 ∫ 1 + x 2 x 2 d x = 2 x 2 tan − 1 x − 2 1 ∫ ( 1 + x 2 x 2 + 1 − 1 + x 2 1 ) d x = 2 x 2 tan − 1 x − 2 1 ∫ ( 1 − 1 + x 2 1 ) d x = 2 x 2 tan − 1 x − 2 1 ( x − tan − 1 x ) + C = 2 x 2 tan − 1 x − 2 x + 2 1 tan − 1 x + C
∫ x cos − 1 x d x
Sol. Let I = ∫ x cos − 1 x dx
put x = cos t , dx = − sin tdt I = − ∫ cos t cos − 1 ( cos t ) sin t d t = − ∫ t sin t cos t d t = − 2 1 ∫ t ( 2 sin t cos t ) d t = − 2 1 ∫ t ( sin 2 t ) d t
Taking t as first function and sin 2 t as second function and integrating by parts, we obtain
= − 2 1 [ t ∫ sin 2 tdt − ∫ ( dt d ( t ) ∫ sin 2 tdt ) dt ] = − 2 1 [ − t 2 cos 2 t + ∫ 2 cos 2 t dt ] = − 2 1 [ − t 2 cos 2 t + 4 sin 2 t ] + C = 4 t cos 2 t − 8 sin 2 t + C = 4 t ( 2 cos 2 t − 1 ) − 8 1 2 sin t cos t + C = 4 t ( 2 cos 2 t − 1 ) − 4 1 cos t 1 − cos 2 t + C = 4 cos − 1 x ( 2 x 2 − 1 ) − 4 1 x 1 − x 2 + C = 4 ( 2 x 2 − 1 ) cos − 1 x − 4 x 1 − x 2 + C
∫ ( sin − 1 x ) 2 dx Sol. Let I = ∫ ( sin − 1 x ) 2 ⋅ 1 dx
Taking ( sin − 1 x ) 2 as first function and 1 as second function and integrating by parts, we obtain
I = ( sin − 1 x ) 2 ∫ 1 d x − ∫ { d x d ( sin − 1 x ) 2 ⋅ ∫ 1 ⋅ d x } d x = x ( sin − 1 x ) 2 − 2 ∫ 1 − x 2 x sin − 1 x d x = x ( sin − 1 x ) 2 − 2 ∫ t ⋅ sin t d t ( 1 − x 2 1 d x = d t Put sin − 1 x = t ⇒ x = sin t )
Again integrating by parts, we obtain
I I = x ( sin − 1 x ) 2 − 2 [ − t cos t − ∫ ( − cos t ) dt ] = x ( sin − 1 x ) 2 + 2 t cos t − 2 sin t + C = x ( sin − 1 x ) 2 + 2 t 1 − sin 2 t − 2 sin t + C = x ( sin − 1 x ) 2 + 2 sin − 1 x 1 − x 2 − 2 x + C
∫ 1 − x 2 x c o s − 1 x dx Sol. Let I = ∫ 1 − x 2 x c o s − 1 x dx
Put cos − 1 x = t ⇒ x = cos t
⇒ 1 − x 2 1 dx = − dt
⇒ I ⇒ I = ∫ t ⋅ cos t d t (Using interagration by parts) = − [ t sin t − ∫ sin t d t ] = − [ t sin t + cos t ] + C = − [ t 1 − cos 2 t + cos t ] + C = − [ 1 − x 2 cos − 1 x + x ] + C
∫ x sec 2 x d x Sol. Let I = ∫ x sec 2 x dx
Taking x as first function and sec 2 x as second function and integrating by parts, we obtain
I = x ∫ sec 2 x d x − ∫ { { d x d ( x ) } ∫ sec 2 x d x } d x = x tan x − ∫ 1 ⋅ tan x d x = x tan x + log ∣ cos x ∣ + C
∫ tan − 1 x d x Sol. Let I = ∫ 1 ⋅ tan − 1 x dx
Taking tan − 1 x as first function and 1 as second function and integrating by parts, we obtain
I = tan − 1 x ∫ 1 d x − ∫ { ( d x d ( tan − 1 x ) ) ∫ 1 ⋅ d x } d x = tan − 1 x ⋅ x − ∫ 1 + x 2 1 ⋅ x d x = x tan − 1 x − 2 1 ∫ 1 + x 2 2 x d x
Let 1 + x 2 = t ⇒ 2 xdx = dt
⇒ I = xtan − 1 x − 2 1 ∫ t 1 dt = x tan − 1 x − 2 1 log ∣ t ∣ + C = x tan − 1 x − 2 1 log 1 + x 2 + C
∫ x ( log x ) 2 d x
Sol. Let I = ∫ x ( log x ) 2 dx
Taking ( log x ) 2 as first function and x as second function and integrating by parts, we obtain I = ( log x ) 2 ∫ x d x − ∫ { ( d x d ( log x ) 2 ) ∫ x d x } d x = 2 x 2 ( log x ) 2 − ∫ [ 2 log x ⋅ x 1 ⋅ 2 x 2 ] d x = 2 x 2 ( log x ) 2 − ∫ x log x d x
Again integrating by parts, we obtain
I = 2 x 2 ( log x ) 2 − [ log x ∫ x d x − ∫ { ( d x d ( log x ) ) ∫ x d x } d x ] = 2 x 2 ( log x ) 2 − [ 2 x 2 log x − ∫ x 1 ⋅ 2 x 2 d x ] = 2 x 2 ( log x ) 2 − 2 x 2 log x + 2 1 ∫ x d x = 2 x 2 ( log x ) 2 − 2 x 2 log x + 4 x 2 + C
∫ ( x 2 + 1 ) log x dx
Sol. Let I = ∫ ( x 2 + 1 ) log x dx : = ∫ x 2 log x d x + ∫ log x d x
Let I = I 1 + I 2
Where, I 1 = ∫ x 2 log x dx and I 2 = ∫ log x dx
I 1 = ∫ x 2 log x d x
Taking log x as first function and x 2 as second function and integrating by parts, we obtain
I 1 I 2 = log x ⋅ ∫ x 2 d x − ∫ { ( d x d ( log x ) ) ∫ x 2 d x } d x = log x ⋅ 3 x 3 − ∫ x 1 ⋅ 3 x 3 d x = 3 x 3 log x − 3 1 ( ∫ x 2 d x ) = 3 x 3 log x − 9 x 3 + C 1 = ∫ log x d x
Taking log x as first function and 1 as second function and integrating by parts, we obtain
I 2 = log x ∫ 1 ⋅ d x − ∫ { ( d x d ( log x ) ) ∫ 1 ⋅ d x } d x = log x ⋅ x − ∫ x 1 ⋅ x d x = x log x − ∫ 1 d x = x log x − x + C 2
Using equations (2) and (3) in (1), we obtain
I = 3 x 3 log x − 9 x 3 + C 1 + x log x − x + C 2 = 3 x 3 log x − 9 x 3 + x log x − x + ( C 1 + C 2 ) = ( 3 x 3 + x ) log x − 9 x 3 − x + C ( ∵ C 1 + C 2 = C )
∫ e x ( sin x + cos x ) dx
Sol. Let I = ∫ e x ( sin x + cos x ) dx
Let f ( x ) = sin x ∴ f ′ ( x ) = cos x
⇒ I = ∫ e x { f ( x ) + f ′ ( x ) } dx It is known that,
∫ e x { f ( x ) + f ′ ( x ) } dx = e x f ( x ) + C
∴ I = e x sin x + C
∫ ( 1 + x ) 2 xe x dx
Sol. Let I = ∫ ( 1 + x ) 2 xe x dx = ∫ e x { ( 1 + x ) 2 x } dx = ∫ e x { ( 1 + x ) 2 1 + x − 1 } d x = ∫ e x { 1 + x 1 − ( 1 + x ) 2 1 } d x
Let f ( x ) = 1 + x 1 ⇒ f ′ ( x ) = ( 1 + x ) 2 − 1
⇒ ∫ ( 1 + x ) 2 xe x dx = ∫ e x { f ( x ) + f ′ ( x ) } dx
It is known that,
∫ e x { f ( x ) + f ′ ( x ) } dx = e x f ( x ) + C
∴ ∫ ( 1 + x ) 2 xe x dx = 1 + x e x + C
∫ e x ( 1 + c o s x 1 + s i n x ) dx
Sol. e x ( 1 + cos x 1 + sin x ) = e x ( 2 cos 2 2 x sin 2 2 x + cos 2 2 x + 2 sin 2 x cos 2 x ) = 2 cos 2 2 x e x ( sin 2 x + cos 2 x ) 2 = 2 1 e x ⋅ ( cos 2 x sin 2 x + cos 2 x ) 2 = 2 1 e x [ tan 2 x + 1 ] 2 = 2 1 e x [ 1 + tan 2 x ] 2 = 2 1 e x [ 1 + tan 2 2 x + 2 tan 2 x ] = 2 1 e x [ sec 2 2 x + 2 tan 2 x ] ∫ ( 1 + cos x ) e x ( 1 + sin x ) d x = ∫ e x [ tan 2 x + 2 1 sec 2 d x ] d x
Let f ( x ) = tan 2 x ⇒ f ′ ( x ) = 2 1 sec 2 2 x
It is known that,
∫ e x { f ( x ) + f ′ ( x ) } dx = e x f ( x ) + C
From equation (1), we obtain ,
∫ ( 1 + c o s x ) e x ( 1 + s i n x ) d x = e x tan 2 x + C
∫ e x ( x 1 − x 2 1 ) dx
Sol. Let I = ∫ e x [ x 1 − x 2 1 ] dx Also, let f ( x ) = x 1 ⇒ f ′ ( x ) = x 2 − 1
It is known that,
∫ e x { f ( x ) + f ′ ( x ) } dx = e x f ( x ) + C
∴ I = x e x + C
∫ ( x − 1 ) 3 ( x − 3 ) e x d x
Sol. Let I = ∫ e x { ( x − 1 ) 3 ( x − 3 ) } dx = ∫ e x { ( x − 1 ) 3 x − 1 − 2 } dx = ∫ e x { ( x − 1 ) 2 1 − ( x − 1 ) 3 2 } dx
Let f ( x ) = ( x − 1 ) 2 1 ⇒ f ′ ( x ) = ( x − 1 ) 3 − 2
It is known that,
∴ ∫ e x { f ( x ) + f ′ ( x ) } d x = e x f ( x ) + C ∫ e x { ( x − 1 ) 3 ( x − 3 ) } d x = ( x − 1 ) 2 e x + C o
∫ e 2 x sin x dx
Sol. Let I = ∫ e 2 x sin x dx
Integrating by parts, we obtain. I = sin x ∫ e 2 x d x − ∫ { ( d x d ( sin x ) ) ∫ e 2 x d x } d x
⇒ I = sin x ⋅ 2 e 2 x − ∫ cos x ⋅ 2 e 2 x dx
⇒ I = 2 e 2 x s i n x − 2 1 ∫ e 2 x cos x dx
Again integrating by parts, we obtain
I = 2 e 2 x sin x − 2 1 [ cos x ∫ e 2 x d x − ∫ { ( d x d ( cos x ) ) ∫ e 2 x d x } d x ]
⇒ I = 2 e 2 x s i n x − 2 1 [ cos x ⋅ 2 e 2 x − ∫ ( − sin x ) 2 e 2 x dx ]
⇒ I = 2 e 2 x ⋅ s i n x − 2 1 [ 2 e 2 x c o s x + 2 1 ∫ e 2 x sin x dx ]
⇒ I = 2 e 2 x s i n x − 4 e 2 x c o s x − 4 1 I
⇒ I + 4 1 I = 2 e 2 x ⋅ s i n x − 4 e 2 x c o s x
⇒ 4 5 I = 2 e 2 x s i n x − 4 e 2 x c o s x
⇒ I = 5 4 [ 2 e 2 x s i n x − 4 e 2 x c o s x ] + C
⇒ I = 5 e 2 x [ 2 sin x − cos x ] + C
∫ sin − 1 ( 1 + x 2 2 x ) dx Sol. Let x = tan θ ⇒ dx = sec 2 θ d θ
∴ sin − 1 ( 1 + x 2 2 x ) = sin − 1 ( 1 + tan 2 θ 2 tan θ ) = sin − 1 ( sin 2 θ ) = 2 θ
⇒ I = ∫ sin − 1 ( 1 + x 2 2 x ) dx = ∫ 2 θ ⋅ sec 2 θ d θ = 2 ∫ θ ⋅ sec 2 θ d θ
Integrating by parts, we obtain
I = 2 [ θ ⋅ ∫ sec 2 θ d θ − ∫ { ( d θ d ( θ ) ) ∫ sec 2 θ d θ } d θ ] = 2 [ θ ⋅ tan θ − ∫ tan θ d θ ] = 2 [ θ tan θ + log ∣ cos θ ∣ + C ] = 2 [ xtan − 1 x + log 1 + x 2 1 + C ] = 2 x tan − 1 x + 2 log ( 1 + x 2 ) − 2 1 + C = 2 x tan − 1 x + 2 [ − 2 1 log ( 1 + x 2 ) ] + C = 2 x tan − 1 x − log ( 1 + x 2 ) + C
Choose the correct answer in Exercises 23 and 24.
∫ x 2 e x 3 dx equals
(A) 3 1 e x 3 + C
(B) 3 1 e x 2 + C
(C) 2 1 e x 3 + C
(D) 2 1 e x 2 + C Sol. (A) Let I = ∫ x 2 e x 3 dx
Put x 3 = t ⇒ 3 x 2 dx = dt
⇒ I = 3 1 ∫ e t d t = 3 1 ( e t ) + C = 3 1 e x 3 + C
Hence, the correct Answer is (A).
∫ e x sec x ( 1 + tan x ) dx
(A) e x cos x + C
(B) e x sec x + C
(C) e x sin x + C
(D) e x tan x + C Sol. (B) Let
I = ∫ e x sec x ( 1 + tan x ) dx = ∫ e x ( sec x + sec x tan x ) dx
Also, if sec x = f ( x ) ⇒ sec x tan x = f ′ ( x )
It is known that, ∫ e x { f ( x ) + f ′ ( x ) } dx = e x f ( x ) + C
∴ I = e x sec x + C
Hence, the correct Answer is (B).
EXERCISE 7.7
Integrate the functions in Exercises 1 to 9.
∫ 4 − x 2 dx Sol. Let I = ∫ 4 − x 2 dx = ∫ ( 2 ) 2 − ( x ) 2 dx
It is known that,
∴ ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 ( a x ) + C I = 2 x 4 − x 2 + 2 4 sin − 1 ( 2 x ) + C = 2 x 4 − x 2 + 2 sin − 1 ( 2 x ) + C
∫ 1 − 4 x 2 dx Sol. Let I = ∫ 1 − 4 x 2 dx = ∫ ( 1 ) 2 − ( 2 x ) 2 dx ,
Let 2 x = t ⇒ 2 dx = dt
∴ I = 2 1 ∫ ( 1 ) 2 − ( t ) 2 dt
It is known that,
{ ∵ ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 ( a x ) + C }
⇒ I = 2 1 [ 2 t 1 − t 2 + 2 1 sin − 1 t ] + C = 4 t 1 − t 2 + 4 1 sin − 1 t + C = 4 2 x 1 − 4 x 2 + 4 1 sin − 1 ( 2 x ) + C = 2 x 1 − 4 x 2 + 4 1 sin − 1 ( 2 x ) + C
∫ x 2 + 4 x + 6 dx
Sol. Let I = ∫ x 2 + 4 x + 6 dx = ∫ x 2 + 4 x + 4 + 2 d x = ∫ ( x 2 + 4 x + 4 ) + 2 d x = ∫ ( x + 2 ) 2 + ( 2 ) 2 d x
It is known that,
{ ∵ ∫ x 2 + a 2 d x = 2 x x 2 + a 2 + 2 a 2 log x + x 2 + a 2 + C } I = 2 ( x + 2 ) x 2 + 4 x + 6 + 2 2 log ( x + 2 ) + x 2 + 4 x + 6 + C = 2 ( x + 2 ) x 2 + 4 x + 6 + log ( x + 2 ) + x 2 + 4 x + 6 + C
∫ x 2 + 4 x + 1 d x
Sol. Let I = ∫ x 2 + 4 x + 1 dx = ∫ ( x 2 + 4 x + 4 ) − 3 d x = ∫ ( x + 2 ) 2 − ( 3 ) 2 d x
It is known that,
{ ∵ ∫ x 2 − a 2 d x = 2 x x 2 − a 2 − 2 a 2 log x + x 2 − a 2 + C } ∴ I = 2 ( x + 2 ) x 2 + 4 x + 1 − 2 3 log ( x + 2 ) + x 2 + 4 x + 1 + C
∫ 1 − 4 x − x 2 dx
Sol. Let I = ∫ 1 − 4 x − x 2 dx = ∫ 1 − ( x 2 + 4 x + 4 − 4 ) d x = ∫ 1 + 4 − ( x + 2 ) 2 d x = ∫ ( 5 ) 2 − ( x + 2 ) 2 d x
It is known that,
∴ ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 ( a x ) + C I = 2 ( x + 2 ) 1 − 4 x − x 2 + 2 5 sin − 1 ( 5 x + 2 ) + C
∫ x 2 + 4 x − 5 dx
Sol. Let I = ∫ x 2 + 4 x − 5 dx = ∫ ( x 2 + 4 x + 4 − 9 ) d x = ∫ ( x + 2 ) 2 − ( 3 ) 2 d x
It is known that,
∫ x 2 − a 2 d x = 2 x x 2 − a 2 − 2 a 2 log x + x 2 − a 2 + C I = 2 ( x + 2 ) x 2 + 4 x − 5 − 2 9 log ( x + 2 ) + x 2 + 4 x − 5 + C
∫ 1 + 3 x − x 2 dx
Sol. Let I = ∫ 1 + 3 x − x 2 dx = ∫ 1 − ( x 2 − 3 x + 4 9 − 4 9 ) d x = ∫ ( 1 + 4 9 ) − ( x − 2 3 ) 2 d x = ∫ ( 2 13 ) 2 − ( x − 2 3 ) 2 d x
It is known that,
{ ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 ( a x ) + C } ∴ I = 2 ( x − 2 3 ) 1 + 3 x − x 2 + 4 × 2 13 sin − 1 ( 2 13 x − 2 3 ) + C = 4 2 x − 3 1 + 3 x − x 2 + 8 13 sin − 1 ( 13 2 x − 3 ) + C
∫ x 2 + 3 x dx
Sol. Let I = ∫ x 2 + 3 x dx = ∫ x 2 + 3 x + 4 9 − 4 9 d x = ∫ ( x + 2 3 ) 2 − ( 2 3 ) 2 d x
It is known that,
{ ∵ ∫ x 2 − a 2 d x = 2 x x 2 − a 2 − 2 a 2 log x + x 2 − a 2 + C } I = 2 ( x + 2 3 ) x 2 + 3 x − 2 4 log ( x + 2 3 ) + x 2 + 3 x + C = 4 ( 2 x + 3 ) x 2 + 3 x − 8 9 log ( x + 2 3 ) + x 2 + 3 x + C
∫ 1 + 9 x 2 dx
Sol. Let I = ∫ 1 + 9 x 2 d x = 3 1 ∫ 9 + x 2 d x = 3 1 ∫ ( 3 ) 2 + x 2 d x
It is known that,
∫ x 2 + a 2 d x = 2 x x 2 + a 2 + 2 a 2 log x + x 2 + a 2 + C
∴ I = 3 1 [ 2 x x 2 + 9 + 2 9 log x + x 2 + 9 ] + C = 6 x x 2 + 9 + 2 3 log x + x 2 + 9 + C
Choose the correct answer in Exercises 10 & 11.
∫ 1 + x 2 dx is equal to ?
(A) 2 x 1 + x 2 + 2 1 log x + 1 + x 2 + C
(B) 3 2 ( 1 + x 2 ) 2 3 + C
(C) 3 2 x ( 1 + x 2 ) 2 3 + C
(D) 2 x 2 1 + x 2 + 2 1 x 2 log x + 1 + x 2 + C Sol. It is known that,
∫ a 2 + x 2 d x = 2 x a 2 + x 2 + 2 a 2 log x + x 2 + a 2 + C ∫ 1 + x 2 d x = 2 x 1 + x 2 + 2 1 log x + 1 + x 2 + C
Hence, the correct Answer is (A).
∫ x 2 − 8 x + 7 d x is equal to ?
(A) 2 1 ( x − 4 ) x 2 − 8 x + 7 + 9 log x − 4 + x 2 − 8 x + 7 + C
(B) 2 1 ( x + 4 ) x 2 − 8 x + 7 + 9 log x + 4 + x 2 − 8 x + 7 + C
(C) 2 1 ( x − 4 ) x 2 − 8 x + 7 − 3 2 log x − 4 + x 2 − 8 x + 7 + C
(D) 2 1 ( x − 4 ) x 2 − 8 x + 7 − 2 9 log x − 4 + x 2 − 8 x + 7 + C
Sol. Let I = ∫ x 2 − 8 x + 7 dx = ∫ ( x 2 − 8 x + 16 ) − 9 d x = ∫ ( x − 4 ) 2 − ( 3 ) 2 d x
It is known that, ∫ x 2 − a 2 dx
= 2 x x 2 − a 2 − 2 a 2 log x + x 2 − a 2 + C
∴ I = 2 ( x − 4 ) x 2 − 8 x + 7 − 2 9 log ( x − 4 ) + x 2 − 8 x + 7 + C
,
Hence, the correct Answer is (D).
EXERCISE 7.8 Evaluate the definite integrals in Exercise 1 to 20.
∫ − 1 1 ( x + 1 ) dx
Sol. Let I = ∫ − 1 1 ( x + 1 ) dx = ( 2 x 2 + x ) − 1 1 = ( 2 1 + 1 ) − ( 2 1 − 1 ) = 2 1 + 1 − 2 1 + 1 = 2
∫ 2 3 x 1 dx
Sol. Let I = ∫ 2 3 x 1 dx = [ log ∣ x ∣ ] 2 3 = log ∣3∣ − log ∣2∣ = log 2 3
∫ 1 2 ( 4 x 3 − 5 x 2 + 6 x + 9 ) d x
Sol. Let I = ∫ 1 2 ( 4 x 3 − 5 x 2 + 6 x + 9 ) dx ∫ 1 2 I ( 4 x 3 − 5 x 2 + 6 x + 9 ) d x = [ 4 ( 4 x 4 ) − 5 ( 3 x 3 ) + 6 ( 2 x 2 ) + 9 ( x ) ] 1 2 = { 2 4 − 3 5 ⋅ ( 2 ) 3 + 3 ( 2 ) 2 + 9 ( 2 ) } − { ( 1 ) 4 − 3 5 ( 1 ) 3 + 3 ( 1 ) 2 + 9 ( 1 ) } = ( 16 − 3 40 + 12 + 18 ) − ( 1 − 3 5 + 3 + 9 ) = 16 − 3 40 + 12 + 18 − 1 + 3 5 − 3 − 9 = 33 − 3 35 = 3 99 − 35 = 3 64
∫ 0 4 π sin 2 xdx
Sol. Let I = ∫ 0 4 π sin 2 x dx = ( 2 − c o s 2 x ) 0 4 π = − 2 1 [ cos 2 ( 4 π ) − cos 0 ] = − 2 1 [ cos ( 2 π ) − cos 0 ] = − 2 1 [ 0 − 1 ] = 2 1
∫ 0 2 π cos 2 x d x
Sol. Let I = ∫ 0 2 π cos 2 x dx = ( 2 s i n 2 x ) 0 2 π = 2 1 [ sin 2 ( 2 π ) − sin 0 ] = 2 1 [ sin π − sin 0 ] = 2 1 [ 0 − 0 ] = 0
∫ 4 5 e x dx
Sol. Let I = ∫ 4 5 e x dx = ( e x ) 4 5 = e 5 − e 4 = e 4 ( e − 1 )
∫ 0 4 π tan x dx
Sol. Let I = ∫ 0 4 π tan x dx = [ − log ∣ cos x ∣ ] 0 4 π = − log cos 4 π + log ∣ cos 0∣ = − log 2 1 + log ∣1∣ { ∵ log ( 1 ) = 0 } = − log ( 2 ) − 2 1 = 2 1 log 2
∫ 6 π 4 π cosec x dx
Sol. Let I = ∫ 6 π 4 π cosec x dx = [ log ∣ cosec x − cot x ∣ ] 6 π 4 π = log cosec 4 π − cot 4 π − log cosec 6 π − cot 6 π = log ∣ 2 − 1∣ − log ∣2 − 3 ∣ = log ( 2 − 3 2 − 1 )
∫ 0 1 1 − x 2 dx
Sol. Let I = ∫ 0 1 1 − x 2 dx = [ sin − 1 x ] 0 1 = sin − 1 ( 1 ) − sin − 1 ( 0 ) = 2 π − 0 = 2 π
∫ 0 1 1 + x 2 dx
Sol. Let I = ∫ 0 1 1 + x 2 dx = [ tan − 1 x ] 0 1 = tan − 1 ( 1 ) − tan − 1 ( 0 ) = 4 π
∫ 2 3 x 2 − 1 dx
Sol. Let I = ∫ 2 3 x 2 − 1 dx = [ 2 1 log x + 1 x − 1 ] 2 3 = 2 1 [ log 3 + 1 3 − 1 − log 2 + 1 2 − 1 ] = 2 1 [ log 4 2 − log 3 1 ] = 2 1 [ log 2 1 − log 3 1 ] = 2 1 [ log 2 3 ]
∫ 0 2 π cos 2 x dx
Sol. Let I = ∫ 0 2 π cos 2 x dx = ∫ 0 2 π ( 2 1 + cos 2 x ) d x = 2 1 ( x + 2 sin 2 x ) 0 2 π = 2 1 [ ( 2 π + 2 sin π ) − ( 0 + 2 sin 0 ) ] = 2 1 [ 2 π + 0 − 0 − 0 ] = 4 π
∫ 2 3 x 2 + 1 xdx
Sol. Let I = ∫ 2 3 x 2 + 1 x dx = 2 1 ∫ 2 3 x 2 + 1 2 x dx = 2 1 [ log ( 1 + x 2 ) ] 2 3 { ∵ x 2 + 1 = t , 2 xdx = dt } = 2 1 [ log ( 1 + ( 3 ) 2 ) − log ( 1 + ( 2 ) 2 ) ] = 2 1 [ log ( 10 ) − log ( 5 )] = 2 1 log ( 5 10 ) = 2 1 log 2
∫ 0 1 5 x 2 + 1 2 x + 3 dx
Sol. Let I = ∫ 0 1 5 x 2 + 1 2 x + 3 dx = 5 1 ∫ 0 1 5 x 2 + 1 5 ( 2 x + 3 ) dx = 5 1 ∫ 0 1 5 x 2 + 1 ( 10 x + 15 ) dx = 5 1 ∫ 0 1 5 x 2 + 1 10 x dx + 3 ∫ 0 1 5 x 2 + 1 1 dx = 5 1 ∫ 0 1 5 x 2 + 1 10 x dx + 3 ∫ 0 1 5 ( x 2 + ( 5 1 ) 2 ) 1 dx
∫ 0 1 xe x 2 dx
Sol. Let I = ∫ 0 1 xe x 2 dx
Put x 2 = t ⇒ 2 x dx = dt
As x → 0 , t → 0 and as x → 1 , t → 1
∴ I = 2 1 ∫ 0 1 e t dt = 2 1 ( e t ) 0 1 = 2 1 e − 2 1 e 0 = 2 1 ( e − 1 )
∫ 1 2 x 2 + 4 x + 3 5 x 2 dx
Sol. Let I = ∫ 1 2 x 2 + 4 x + 3 5 x 2 dx
Dividing 5 x 2 by x 2 + 4 x + 3 , we obtain I = ∫ 1 2 { 5 − x 2 + 4 x + 3 20 x + 15 } d x = ∫ 1 2 5 d x − ∫ 1 2 x 2 + 4 x + 3 20 x + 15 d x = [ 5 x ] 1 2 − ∫ 1 2 x 2 + 4 x + 3 20 x + 15 d x
⇒ I I 1 = 5 − I 1 , where = ∫ 1 2 x 2 + 4 x + 3 20 x + 15 dx
Consider I 1 = ∫ 1 2 x 2 + 4 x + 3 20 x + 15 dx
Let 20 x + 15 = A d x d ( x 2 + 4 x + 3 ) + B = 2 A x + ( 4 A + B )
Equating the coefficients of x and constant term, we obtain
⇒ A = 10 and B = − 25 I 1 = 10 ∫ 1 2 x 2 + 4 x + 3 2 x + 4 dx − 25 ∫ 1 2 x 2 + 4 x + 3 dx
⇒ ⇒ ( 2 x + 4 ) dx = dt I 1 = 10 ∫ 8 15 t dt − 25 ∫ 1 2 ( x + 2 ) 2 − 1 2 dx = 10 [ log t ] 8 15 − 25 [ 2 1 log ( x + 2 + 1 x + 2 − 1 ) ] 1 2 = [ 10 log 15 − 10 log 8 ] − 25 [ 2 1 log 5 3 − 2 1 log 4 2 ] = [ 10 log ( 5 × 3 ) − 10 log ( 4 × 2 )] − 2 25 [ log 3 − log 5 − log 2 + log 4 ] = [ 10 log 5 + 10 log 3 − 10 log 4 − 10 log 2 ] − 2 25 [ log 3 − log 5 − log 2 + log 4 ]
= [ 10 + 2 25 ] log 5 + [ − 10 − 2 25 ] log 4 + [ 10 − 2 25 ] log 3 + [ − 10 + 2 25 ] log 2 = 2 45 log 5 − 2 45 log 4 − 2 5 log 3 + 2 5 log 2 = 2 45 log 4 5 − 2 5 log 2 3
Substituting the value of I 1 in (1), we obtain
I = 5 − [ 2 45 log 4 5 − 2 5 log 2 3 ] = 5 − 2 5 [ 9 log 4 5 − log 2 3 ]
∫ 0 4 π ( 2 sec 2 x + x 3 + 2 ) dx Sol. Let I = ∫ 0 4 π ( 2 sec 2 x + x 3 + 2 ) dx
= ( 2 tan x + 4 x 4 + 2 x ) 0 4 π = { ( 2 tan 4 π + 4 1 ( 4 π ) 4 + 2 ( 4 π ) ) − ( 2 tan 0 + 0 + 0 )} = 2 tan 4 π + 4 5 π 4 + 2 π = 2 + 2 π + 1024 π 4
∫ 0 π ( sin 2 2 x − cos 2 2 x ) dx Sol. Let I = ∫ 0 π ( sin 2 2 x − cos 2 2 x ) dx
= − ∫ 0 π ( cos 2 2 x − sin 2 2 x ) d x = − ∫ 0 π cos x d x = ( − sin x ) 0 π = − [ sin π − sin 0 ] = 0
∫ 0 2 x 2 + 4 6 x + 3 dx Sol. Let I = ∫ 0 2 x 2 + 4 6 x + 3 dx = 3 ∫ 0 2 x 2 + 4 2 x + 1 dx
= 3 ∫ 0 2 x 2 + 4 2 x d x + 3 ∫ 0 2 x 2 + 2 2 1 d x { ∵ ∫ a 2 + x 2 d x = a 1 tan − 1 a x } = 3 [ log ( x 2 + 4 ) ] 0 2 + 2 3 ( tan − 1 2 x ) 0 2 = { 3 log ( 2 2 + 4 ) + 2 3 tan − 1 ( 2 2 ) } − { 3 log ( 0 + 4 ) + 2 3 tan − 1 ( 2 0 ) } = 3 log 8 + 2 3 tan − 1 1 − 3 log 4 − 2 3 tan − 1 0 = 3 log 8 + 2 3 ( 4 π ) − 3 log 4 − 0 = 3 log ( 4 8 ) + 8 3 π = 3 log 2 + 8 3 π
∫ 0 1 ( xe x + sin 4 π x ) dx Sol. Let I = ∫ 0 1 ( xe x + sin 4 π x ) dx
= ( x e x ) 0 1 − ∫ 0 1 e x d x − [ π 4 cos 4 π x ] 0 1 = [ x e x − e x − π 4 cos 4 π x ] 0 1 = ( 1 ⋅ e 1 − e 1 − π 4 cos 4 π ) − ( 0 ⋅ e 0 − e 0 − π 4 cos 0 ) = e − e − π 4 ( 2 1 ) + 1 + π 4 = 1 + π 4 − π 2 2
Choose the correct answer in Exercises 21 & 22.
∫ 1 3 1 + x 2 dx equals
(A) 3 π
(B) 3 2 π
(C) 6 π
(D) 12 π Sol. (D) I = ∫ 1 3 1 + x 2 dx = ( tan − 1 x ) 1 3
= tan − 1 3 − tan − 1 1 = 3 π − 4 π = 12 π .
Hence, the correct Answer is (D).
∫ 0 3 2 4 + 9 x 2 dx equals?
(A) 6 π
(B) 12 π
(C) 24 π
(D) 4 π Sol. (C) I = ∫ 0 3 2 4 + 9 x 2 dx = ∫ 0 3 2 ( 2 ) 2 + ( 3 x ) 2 dx
= 3 1 [ 2 1 tan − 1 2 3 x ] 0 3 2 = 6 1 [ tan − 1 ( 2 3 x ) ] 0 3 2 = 6 1 tan − 1 ( 2 3 ⋅ 3 2 ) − 6 1 tan − 1 0 = 6 1 tan − 1 1 − 0 = 6 1 × 4 π = 24 π .
Hence, the correct Answer is (C).
EXERCISE 7.9 Evaluate the integrals in Q. 1 to 8 using substitution.
∫ 0 1 x 2 + 1 x dx Sol. Let I = ∫ 0 1 x 2 + 1 x dx
Put x 2 + 1 = t ⇒ 2 x dx = dt
When x = 0 , t = 1 and when x = 1 , t = 2
∴ ∫ 0 1 x 2 + 1 x dx = 2 1 ∫ 1 2 t dt = 2 1 [ log ∣ t ∣ ] 1 2 = 2 1 [ log 2 − log 1 ] = 2 1 log 2
∫ 0 2 π sin ϕ cos 5 ϕ d ϕ
Sol. Let I = ∫ 0 2 π sin ϕ cos 5 ϕ d ϕ = ∫ 0 2 π sin ϕ cos 4 ϕ cos ϕ d ϕ = ∫ 0 π /2 sin ϕ ( 1 − sin 2 ϕ ) 2 ⋅ cos ϕ d ϕ { ∵ cos 2 x = 1 − sin 2 x }
Put sin ϕ = t ⇒ cos ϕ d ϕ = dt
When ϕ = 0 , t = 0 and when ϕ = 2 π , t = 1
∴ I = ∫ 0 1 t ( 1 − t 2 ) 2 dt = ∫ 0 1 t 2 1 ( 1 + t 4 − 2 t 2 ) dt
= ∫ 0 1 [ t 2 1 + t 2 9 − 2 t 2 5 ] dt = [ 2 3 t 2 3 + 2 11 t 2 11 − 2 7 2 t 2 7 ] 0 1 = 3 2 + 11 2 − 7 4 = 231 154 + 42 − 132 = 231 64
∫ 0 1 sin − 1 ( 1 + x 2 2 x ) dx
Sol. Let I = ∫ 0 1 sin − 1 ( 1 + x 2 2 x ) dx Put x = tan θ ⇒ dx = sec 2 θ d θ
When x = 0 , θ = 0 and when x = 1 , θ = 4 π
I = ∫ 0 4 π sin − 1 ( 1 + t a n 2 θ 2 t a n θ ) sec 2 θ d θ
⇒ I = ∫ 0 4 π sin − 1 ( sin 2 θ ) sec 2 θ d θ
I = ∫ 0 4 π 2 θ ⋅ sec 2 θ d θ = 2 ∫ 0 4 π θ ⋅ sec 2 θ d θ
Taking θ as first function and sec 2 θ as second function and integrating by parts, we obtain
I = 2 [ θ ∫ sec 2 θ d θ − ∫ { ( d θ d ( θ ) ) ∫ sec 2 θ d θ } d θ ] 0 4 π
= 2 [ θ tan θ − ∫ tan θ d θ ] 0 π /4 = 2 [ θ tan θ + log ∣ cos θ ∣ ] 0 4 π = 2 [ 4 π tan 4 π + log cos 4 π − log ∣ cos 0∣ ] = 2 [ 4 π + log ( 2 1 ) − log 1 ] = 2 [ 4 π − 2 1 log 2 ] = 2 π − log 2
∫ 0 2 x x + 2 dx
Sol. Let I = ∫ 0 2 x x + 2 dx
Put x + 2 = t 2 ⇒ dx = 2 tdt , when, x = 0 , t = 2 and when x = 2 , t = 2
∴ ∫ 0 2 x x + 2 dx = ∫ 2 2 ( t 2 − 2 ) t 2 2 tdt = 2 ∫ 2 2 ( t 2 − 2 ) t 2 dt = 2 ∫ 2 2 ( t 4 − 2 t 2 ) dt
= 2 [ 5 t 5 − 3 2 t 3 ] 2 2
= 2 [ 5 32 − 3 16 − 5 4 2 + 3 4 2 ]
= 2 [ 15 96 − 80 − 12 2 + 20 2 ]
= 2 [ 15 16 + 8 2 ]
= 15 16 ( 2 + 2 )
= 15 16 2 ( 2 + 1 )
∫ 0 2 π 1 + c o s 2 x s i n x dx
Sol. Let I = ∫ 0 2 π 1 + c o s 2 x s i n x dx
Put cos x = t ⇒ − sin x dx = dt
When x = 0 , t = 1 and when x = 2 π , t = 0 ⇒ ∫ 0 2 π 1 + cos 2 x sin x d x = − ∫ 1 0 1 + t 2 d t = − [ tan − 1 t ] 1 0 = − [ tan − 1 0 − tan − 1 1 ] = − [ − 4 π ] = 4 π
∫ 0 2 x + 4 − x 2 dx
Sol. Let I = ∫ 0 2 x + 4 − x 2 dx = ∫ 0 2 − ( x 2 − x − 4 ) dx = ∫ 0 2 − ( x 2 − x + 4 1 − 4 1 − 4 ) dx = ∫ 0 2 − [ ( x − 2 1 ) 2 − 4 17 ] dx = ∫ 0 2 ( 2 17 ) 2 − ( x − 2 1 ) 2 dx
Put x − 2 1 = t ⇒ dx = dt , when x = 0 , t = − 2 1 and when x = 2 , t = 2 3
∴ ∫ 0 2 ( 2 17 ) 2 − ( x − 2 1 ) 2 dx = ∫ − 2 1 2 3 ( 2 17 ) 2 − t 2 dt = 2 ( 2 17 ) 1 log 2 17 2 17 + t − t − 2 1 2 3
= 17 1 [ log 2 17 − 2 3 2 17 + 2 3 − log 2 17 + 2 17 − = 17 1 [ log 17 − 3 17 + 3 − log 17 + 1 17 − 1 = 17 1 log ( 17 − 3 17 + 3 × 17 − 1 17 + 1 ) = 17 1 log [ 17 + 3 − 4 17 17 + 3 + 4 17 ] = 17 1 log [ 20 − 4 17 20 + 4 17 ] = 17 1 log ( 5 − 17 5 + 17 ) = 17 1 log [ 25 − 17 ( 5 + 17 ) ( 5 + 17 ) ] = 17 1 log [ 8 25 + 17 + 10 17 ] = 17 1 log ( 8 42 + 10 17 ) = 17 1 log ( 4 21 + 5 17 )
∫ − 1 1 x 2 + 2 x + 5 dx
Sol. Let I = ∫ − 1 1 x 2 + 2 x + 5 dx = ∫ − 1 1 ( x 2 + 2 x + 1 ) + 4 dx
Put x + 1 = t ⇒ dx = dt
When x = − 1 , t = 0 and when x = 1 , t = 2 ∫ − 1 1 ( x + 1 ) 2 + ( 2 ) 2 dx = ∫ 0 2 t 2 + 2 2 dt = [ 2 1 tan − 1 2 t ] 0 2 = 2 1 tan − 1 1 − 2 1 tan − 1 0 = 2 1 ( 4 π ) = 8 π
∫ 1 2 ( x 1 − 2 x 2 1 ) e 2 x dx
Sol. Let I = ∫ 1 2 ( x 1 − 2 x 2 1 ) e 2 x dx
Put 2 x = t ⇒ 2 dx = dt
When x = 1 , t = 2 and when x = 2 , t = 4
∴ ∫ 1 2 ( x 1 − 2 x 2 1 ) e 2 x dx = 2 1 ∫ 2 4 ( t 2 − t 2 2 ) e t dt = ∫ 2 4 ( t 1 − t 2 1 ) e t dt [ ∴ ∫ e t [ f ( t ) + f ′ ( t ) ] dt = e t f ( t ) + C ]
= [ t e t ] 2 4 = 4 e 4 = 4 e 2 ( e 2 − 2 )
Choose the correct answer in Exercises 9 & 10.
The value of the integral ∫ 3 1 1 x 4 ( x − x 3 ) 3 1 d x is ?
(A) 6
(B) 0
(C) 3
(D) 4
Sol. (A) Let I = ∫ 3 1 1 x 4 ( x − x 3 ) 3 1 dx ⇒ ⇒ I = ∫ 3 1 1 x 4 x ( x 2 1 − 1 ) 3 1 d x I = ∫ 3 1 1 x 3 ( x 2 1 − 1 ) 3 1 d x Put x 2 1 − 1 = t 3 , − x 3 2 ⋅ d x = 3 t 2 d t , x 3 d x = 2 − 3 t 2 d t
When x = 1 then t = 0 and when x = 3 1 then t = 2
= − 2 3 ∫ 2 0 ( t 3 ) 1/3 ⋅ t 2 dt = 2 3 ∫ 0 2 t 3 ⋅ dt = 2 3 [ 4 t 4 ] 0 2 = 2 3 × 4 1 [ 2 4 − 0 ] = 8 3 × 16 = 6
Hence, the correct Answer is (A)
If f ( x ) = ∫ 0 x t sin t dt , then f ′ ( x ) is
(A) cos x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Sol. (B) f ( x ) = ∫ 0 x t sin t dt
Differentiation on both sides f ′ ( x ) f ′ ( x ) = ( t sin t ) 0 x = x sin x
Hence, the correct Answer is (B).
EXERCISE 7.10 By using the properties of definite integrals, evaluate the in Exercises 1 to 19.
∫ 0 2 π cos 2 x dx
Sol. I = ∫ 0 2 π cos 2 x dx ⇒ I = ∫ 0 2 π cos 2 ( 2 π − x ) dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
Adding (1) and (2), we obtain
2 I = ∫ 0 2 π ( sin 2 x + cos 2 x ) d x
⇒ 2 I = ∫ 0 2 π 1 ⋅ dx ⇒ 2 I = [ x ] 0 2 π
⇒ 2 I = 2 π ⇒ I = 4 π
∫ 0 2 π s i n x + c o s x s i n x dx
Sol. Let I = ∫ 0 2 π s i n x + c o s x s i n x dx I = ∫ 0 2 π sin ( 2 π − x ) + cos ( 2 π − x ) sin ( 2 π − x ) d x ( ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x )
Adding (1) and (2), we obtain
2 I = ∫ 0 2 π s i n x + c o s x s i n x + c o s x d x
⇒ ⇒ 2 I = ∫ 0 2 π 1 ⋅ dx 2 I = 2 π ⇒ ⇒ 2 I = [ x ] 0 2 π I = 4 π
∫ 0 2 π s i n 2 3 x + c o s 2 3 x s i n 2 3 xdx
Sol. Let I = ∫ 0 2 π s i n 2 3 x + c o s 2 3 x s i n 2 3 x dx ⇒ I = ∫ 0 2 π sin 2 3 ( 2 π − x ) + cos 2 3 ( 2 π − x ) sin 2 3 ( 2 π − x ) dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 2 π s i n 2 3 x + c o s 2 3 x c o s 2 3 x dx
Adding (1) and (2), we obtain
2 I = ∫ 0 2 π s i n 2 3 x + c o s 2 3 x s i n 2 3 x + c o s 2 3 x d x
⇒ ⇒ 2 I = ∫ 0 2 π 1 ⋅ dx = [ x ] 0 2 π 2 I = 2 π ⇒ I = 4 π
∫ 0 2 π s i n 5 x + c o s 5 x c o s 5 x dx
Sol. Let I = ∫ 0 2 π s i n 5 x + c o s 5 x c o s 5 x dx ⇒ I = ∫ 0 2 π s i n 5 ( 2 π − x ) + c o s 5 ( 2 π − x ) c o s 5 ( 2 π − x ) dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 2 π s i n 5 x + c o s 5 x s i n 5 x d x
Adding (1) and (2), we obtain
2 I = ∫ 0 2 π s i n 5 x + c o s 5 x s i n 5 x + c o s 5 x d x
⇒ ⇒ ⇒ 2 I = ∫ 0 2 π 1 ⋅ dx 2 I = 2 π I = 4 π ⇒ ⇒
∫ − 5 5 ∣ x + 2∣ dx
Sol. Let I = ∫ − 5 5 ∣ x + 2∣ dx
We know that ∣ x + 2∣ = { ( x + 2 ) − ( x + 2 ) , x ≥ − 2 , x < − 2 I = ∫ − 5 − 2 ∣ x + 2∣ dx + ∫ − 2 5 ∣ x + 2∣ dx
∴ = = I = ∫ − 5 − 2 − ( x + 2 ) d x + ∫ − 2 5 ( x + 2 ) d x ( ∵ ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x ) I = − [ 2 x 2 + 2 x ] − 5 − 2 + [ 2 x 2 + 2 x ] − 2 5 = − [ 2 ( − 2 ) 2 + 2 ( − 2 ) − 2 ( − 5 ) 2 − 2 ( − 5 ) ] + [ 2 ( 5 ) 2 + 2 ( 5 ) − 2 ( − 2 ) 2 − 2 ( − 2 ) ] − [ 2 − 4 − 2 25 + 10 ] + [ 2 25 + 10 − 2 + 4 ] − 2 + 4 + 2 25 − 10 + 2 25 + 10 − 2 + 4 = 29
∫ 2 8 ∣ x − 5∣ dx
Sol. Let I = ∫ 2 8 ∣ x − 5∣ dx
We know that ∣ x − 5∣ = { ( x − 5 ) − ( x − 5 ) , , x ≥ 5 x < 5 I = I = = = = ∫ 2 5 ∣ x − 5∣ dx + ∫ 5 8 ∣ x − 5∣ dx ∫ 2 5 − ( x − 5 ) dx + ∫ 5 8 ( x − 5 ) dx ( ∵ ∫ a b f ( x ) dx = ∫ a c f ( x ) dx + ∫ c b f ( x ) dx ) − [ 2 x 2 − 5 x ] 2 5 + [ 2 x 2 − 5 x ] 5 8 − [ 2 25 − 25 − 2 + 10 ] + [ 32 − 40 − 2 25 + 25 ] 9
∫ 0 1 x ( 1 − x ) n dx
Sol. Let I = ∫ 0 1 x ( 1 − x ) n dx ∴ I = ∫ 0 1 ( 1 − x ) ( 1 − ( 1 − x ) ) n dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx ) = ∫ 0 1 ( 1 − x ) ( x ) n dx = ∫ 0 1 ( x n − x n + 1 ) dx = [ n + 1 x n + 1 − n + 2 x n + 2 ] 0 1 = [ n + 1 1 − n + 2 1 ] = ( n + 1 ) ( n + 2 ) ( n + 2 ) − ( n + 1 ) = ( n + 1 ) ( n + 2 ) 1
∫ 0 4 π log ( 1 + tan x ) dx
Sol. Let I = ∫ 0 4 π log ( 1 + tan x ) dx
∴ I = ∫ 0 4 π log [ 1 + tan ( 4 π − x ) ] dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 4 π log { 1 + 1 + tan 4 π tan x tan 4 π − tan x } dx ⇒ I = ∫ 0 4 π log { 1 + 1 + tan x 1 − tan x } dx ⇒ I = ∫ 0 4 π { log ( 1 + tan x ) 2 } dx ⇒ I = ∫ 0 4 π log 2 dx − ∫ 0 4 π log ( 1 + tan x ) dx ⇒ I = ∫ 0 4 π log 2 dx − I [From (1)] ⇒ 2 I = [ x log 2 ] 0 4 π ⇒ 2 I = 4 π log 2 ⇒ I = 8 π log 2
∫ 0 2 x 2 − x dx
Sol. Let I I = ∫ 0 2 x 2 − x d x = ∫ 0 2 ( 2 − x ) x d x ( ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x ) = ∫ 0 2 { 2 x 2 1 − x 2 3 } d x = [ 2 ( 2 3 x 2 3 ) − 2 5 x 2 5 ] 0 2 = [ 3 4 x 2 3 − 5 2 x 2 5 ] 0 2 = 3 4 ( 2 ) 2 3 − 5 2 ( 2 ) 2 5 = 3 4 × 2 2 − 5 2 × 4 2 = 3 8 2 − 5 8 2 = 15 40 2 − 24 2 = 15 16 2
∫ 0 2 π ( 2 log sin x − log sin 2 x ) dx Sol. Let
I I = ∫ 0 2 π ( 2 log sin x − log sin 2 x ) d x = ∫ 0 2 π log ( sin 2 x sin 2 x ) d x = ∫ 0 2 π log ( 2 sin x cos x sin 2 x ) d x = ∫ 0 2 π log ( 2 tan x ) d x = ∫ 0 2 π log tan x d x − ∫ 0 2 π log 2 d x
⇒ I = I ′ − 2 π log 2
Now,
I ′ = ∫ 0 2 π log tan x d x = ∫ 0 2 π log tan ( 2 π − x ) d x ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx ) = ∫ 0 2 π log cot x dx
Adding equation (2) and (3)
2 I ′ 2 I ′ I ′ I I = ∫ 0 2 π ( log tan x + log cot x ) d x = ∫ 0 2 π ( log tan x cot x ) d x = ∫ 0 2 π log 1 d x = 0 = 0 put in equation ( 1 ) = 0 − 2 π log 2 = 2 π log 2 1
∫ 2 − π 2 π sin 2 x dx
Sol. Let I = ∫ 2 − π 2 π sin 2 x dx ⇒ 2 ∫ 0 2 π sin 2 x dx
∵ ∫ − a a f ( x ) dx = 2 ∫ 0 a f ( x ) dx , when f ( x ) is even function
⇒ I = 2 ∫ 0 2 π sin 2 ( 2 π − x ) dx [ ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx ]
⇒ I = 2 ∫ 0 2 π cos 2 x dx
Adding equation (1) and (2),
2 I = 2 ∫ 0 2 π ( sin 2 x + cos 2 x ) dx = 2 ∫ 0 2 π 1 dx = 2 ( x ) 0 2 π = 2 ⋅ 2 π = π ⇒ 2 I = π ⇒ I = 2 π ⇒ I = 2 π
∫ 0 π 1 + s i n x x dx
Sol. Let I = ∫ 0 π 1 + s i n x x dx
⇒ I = ∫ 0 π 1 + s i n ( π − x ) ( π − x ) dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 π 1 + s i n x ( π − x ) dx
Adding eq. (1) and (2), we obtain ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ 2 I 2 I 2 I 2 I 2 I = π [ tan π − tan 0 ] − π [ sec π − sec 0 ] I = ∫ 0 π 1 + s i n x π dx = π ∫ 0 π ( 1 + s i n x ) ( 1 − s i n x ) ( 1 − s i n x ) dx = π ∫ 0 π c o s 2 x 1 − s i n x dx = π ∫ 0 π { sec 2 x − tan x sec x } dx = π [ tan x − sec x ] 0 π π [ 0 − 0 ] − π [ − 1 − 1 ] = 2 π ⇒ I = π
∫ − 2 π 2 π sin 7 x dx
As sin 7 ( − x ) = ( sin ( − x ) ) 7 = ( − sin x ) 7 = − sin 7 x , therefore, sin 7 x is an odd function.
It is known that, if f ( x ) is an odd function, then ∫ − a a f ( x ) dx = 0 ∴ I = ∫ − 2 π 2 π sin 7 x dx = 0
∫ 0 2 π cos 5 x dx
Sol. Let I = ∫ 0 2 π cos 5 x dx f ( x ) = cos 5 x , then f ( 2 π − x ) = cos 5 ( 2 π − x ) = cos 5 x = f ( x )
I = 2 ∫ 0 π cos 5 ( π − x ) d x = − 2 ∫ 0 π cos 5 x d x ( ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x ) I = − 2 ∫ 0 π cos 5 x d x = − I ⇒ 2 I = 0 ⇒ I = 0
[From eq. (1)]
15. ∫ 0 2 π 1 + s i n x c o s x s i n x − c o s x dx
Sol . Let I = ∫ 0 2 π 1 + s i n x c o s x s i n x − c o s x dx
⇒
I = ∫ 0 2 π 1 + sin ( 2 π − x ) cos ( 2 π − x ) sin ( 2 π − x ) − cos ( 2 π − x ) d x ( ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x )
⇒ I = ∫ 0 2 π 1 + s i n x c o s x c o s x − s i n x d x
Adding (1) and (2), we obtain,
2 I = ∫ 0 2 π 1 + s i n x c o s x 0 d x ⇒ I = 0
∫ 0 π log ( 1 + cos x ) dx
Sol. Let I = ∫ 0 π log ( 1 + cos x ) dx ⇒ I = ∫ 0 π log ( 1 + cos ( π − x )) dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 π log ( 1 − cos x ) dx
Adding (1) and (2), we obtain
2 I = ∫ 0 π { log ( 1 + cos x ) + log ( 1 − cos x )} d x
⇒ 2 I = ∫ 0 π log ( 1 − cos 2 x ) dx
⇒ 2 I = ∫ 0 π log sin 2 x dx
⇒ 2 I = 2 ∫ 0 π log sin x dx
⇒ I = ∫ 0 π log sin x dx
( ∵ sin ( π − x ) = sin x )
∴ I = 2 ∫ 0 2 π log sin x d x
⇒ I = 2 ∫ 0 2 π log sin ( 2 π − x ) dx = 2 ∫ 0 2 π log cos x dx
Adding (4) and (5), we obtain
2 I = 2 ∫ 0 2 π ( log sin x + log cos x ) d x
⇒ I = ∫ 0 2 π ( log sin x + log cos x + log 2 − log 2 ) dx
⇒ I = ∫ 0 2 π ( log 2 sin x cos x − log 2 ) d x
⇒ I = ∫ 0 2 π log sin 2 xdx − ∫ 0 2 π log 2 dx
Let 2 x = t ⇒ 2 dx = dt
When x = 0 , t = 0 and when x = 2 π , t = π
∴ I = 2 1 ∫ 0 π log sin t dt − 2 π log 2
⇒ I = 2 1 ∫ 0 π log sin x dx − 2 π log 2
⇒ I = 2 1 I − 2 π log 2
⇒ 2 I = − 2 π log 2
⇒ I = − π log 2 [ ∵ ∫ a b f ( x ) dx = ∫ a b f ( t ) dt ]
[From eq.(3)]
∫ 0 a x + a − x x dx
Sol. Let I = ∫ 0 a x + a − x x dx
I = ∫ 0 a a − x + x a − x dx
( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx ) Adding eq. (1) and (2), we obtain
2 I = ∫ 0 a x + a − x x + a − x dx ⇒ 2 I = ∫ 0 a 1 dx
⇒ 2 I = [ x ] 0 a ⇒ 2 I = a ⇒ I = 2 a
∫ 0 4 ∣ x − 1∣ dx Sol. Let I = ∫ 0 4 ∣ x − 1∣ dx
we know that ∣ x − 1∣ = { ( x − 1 ) − ( x − 1 ) , , x ≥ 1 x < 1
I = ∫ 0 1 ∣ x − 1∣ dx + ∫ 1 4 ∣ x − 1∣ dx = ∫ 0 1 − ( x − 1 ) dx + ∫ 1 4 ( x − 1 ) dx [ ∵ ∫ a b f ( x ) dx = ∫ a c f ( x ) dx + ∫ c b f ( x ) dx ]
I = − [ 2 x 2 − x ] 0 1 + [ 2 x 2 − x ] 1 4 = − [ ( 2 1 − 1 ) − 0 ] + [ ( 2 16 − 4 ) − ( 2 1 − 1 ) ] = 2 1 + 8 − 4 + 2 1 = 5
Show that ∫ 0 a f ( x ) g ( x ) dx = 2 ∫ 0 a f ( x ) dx , if f and g are defined as f ( x ) = f ( a − x ) and g ( x ) + g ( a − x ) = 4
Sol. Let I = ∫ 0 a f ( x ) g ( x ) dx
⇒ I = ∫ 0 a f ( a − x ) g ( a − x ) dx ( ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 a f ( x ) g ( a − x ) dx
Adding eq. (1) and (2), we obtain
2 I = ∫ 0 a { f ( x ) g ( x ) + f ( x ) g ( a − x )} dx
⇒ 2 I = ∫ 0 a f ( x ) { g ( x ) + g ( a − x )} dx
⇒ 2 I = ∫ 0 a f ( x ) × 4 dx [ Given g ( x ) + g ( a − x ) = 4 ]
⇒ 2 I = 4 ∫ 0 a f ( x ) dx
⇒ I = 2 ∫ 0 a f ( x ) dx
Hence proved
Choose the correct answer in Exercises 20 & 21.
The value of ∫ − 2 π 2 π ( x 3 + x cos x + tan 5 x + 1 ) dx is?
(A) 0
(B) 2
(C) π
(D) 1 Sol. (C)
Let I = ∫ − 2 π 2 π ( x 3 + x cos x + tan 5 x + 1 ) dx
⇒ I = ∫ − 2 π 2 π x 3 dx + ∫ − 2 π 2 π x cos x
+ ∫ − 2 π 2 π tan 5 x d x + ∫ − 2 π 2 π 1 ⋅ d x
It is known that if f ( x ) is an even function, then ∫ − a a f ( x ) dx = 2 ∫ 0 a f ( x ) dx and if is an odd function, then ∫ − a a f ( x ) dx = 0
[ ∵ x 3 , x cos x and tan 5 x are odd functions]
I = [ x ] 2 − π 2 π ⇒ I = [ 2 π + 2 π ] = 2 2 π = π
Hence, the correct Answer (C).
21. The value of ∫ 0 2 π log ( 4 + 3 c o s x 4 + 3 s i n x ) dx is ?
(A) 2
(B) 4 3
(C) 0
(D) -2
Sol. (C)
Let I = ∫ 0 2 π log ( 4 + 3 c o s x 4 + 3 s i n x ) d x
⇒ I = ∫ 0 2 π log [ 4 + 3 c o s ( 2 π − x ) 4 + 3 s i n ( 2 π − x ) ] dx ( ∵ ∫ 0 a f ( x ) dx = ∫ 0 a f ( a − x ) dx )
⇒ I = ∫ 0 2 π log ( 4 + 3 s i n x 4 + 3 c o s x ) d x
Adding (1) and (2), we obtain
2 I = ∫ 0 2 π { log ( 4 + 3 cos x 4 + 3 sin x ) + log ( 4 + 3 sin x 4 + 3 cos x ) } d x 2 I = ∫ 0 2 π { log ( 4 + 3 cos x 4 + 3 sin x × 4 + 3 sin x 4 + 3 cos x ) } d x
⇒ 2 I = ∫ 0 2 π log 1 dx ⇒ 2 I = ∫ 0 2 π 0 dx ⇒ I = 0
Hence, the correct Answer is (C).
3.0 Class 12 Maths NCERT Solutions – Chapter-wise Links Access NCERT Solutions for Class 12 Maths covering all chapters, with clear exercise answers, important formulas, and step-by-step solutions for understanding concepts and solving questions.
4.0 Class 12 Maths Chapter 7 Integrals Exercise-wise Solutions 5.0 Key Features and Benefits of Class 12 Maths Chapter 7 Integrals Step-by-Step Solutions: Detailed solutions help students follow lengthy integration calculations easily and present their answers properly in exams. NCERT-Based Practice: The exercises are aligned with the NCERT syllabus and cover question types that are useful for CBSE board exam preparation. Better Calculation Accuracy: Regular practice with the solutions helps improve accuracy and reduces mistakes while solving lengthy integration problems. Useful for Competitive Exams: A strong understanding of integration can support preparation for Mathematics Olympiads and other competitive examinations. Strong Foundation for Further Topics: Learning the concepts of integration thoroughly makes the next chapter, Applications of Integrals, easier to understand and practise.