Integral

Integral calculus is a fundamental branch of mathematics that extends our understanding of accumulation, areas under curves, and much more. Whether you're diving into basic integrals or exploring the complexities of double and triple integrals, mastering this subject is crucial for success in various scientific and engineering fields. This blog will guide you through the key concepts and applications of integral calculus, providing clear definitions, formulas, solved examples, and practice problems.

1.0Integration Definition

In calculus, an Integral is a tool used to add up tiny pieces to find the total area or volume of a continuous shape. The process of calculating an integral is called integration, and it was once known as quadrature. When we estimate an integral using methods other than exact calculations, it's called numerical integration. There are two main types of integrals: definite integrals, which give a specific number representing the algebraic area or volume between two points, and indefinite integrals, which represent a general formula plus a constant.

2.0Indefinite Integral

The indefinite integral, also known as the antiderivative, represents a family of functions whose derivative is the given function. It includes an arbitrary constant C, since differentiation of a constant is zero. The notation for an indefinite integral is:

$\int f(x)dx=F(x)+C$

where F(x) is the antiderivative of f(x).

3.0Definite Integral

The definite integral calculates the net area under a curve within a specified interval [a, b]. Unlike the indefinite integral, it produces a specific numerical value. The notation for a definite integral is:

${\int }_a^{b}f(x)dx$

This value can be interpreted as the total accumulation of the quantity represented by f(x) from x = a to x = b.

4.0Integration Formulas

• $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C;n\ne -1$
• $\int \frac{dx}{ax+b}=\frac{1}{a}\ln |ax+b|+C$
• $\int e^{ax+b}dx=\frac{1}{a}{e}^{ax+b}+C$  
• $\int a^{px+q}dx=\frac{1}{p}\frac{a^{px+q}}{\ln a}+C,(a>0)$
• $\int \sin (ax+b)dx=-\frac{1}{a}\cos (ax+b)+C$
• $\int \cos (ax+b)dx=\frac{1}{a}\sin (ax+b)+C$
• $\int \tan (ax+b)dx=\frac{1}{a}\ln |\sec (ax+b)|+C$
• $\int \cot (ax+b)dx=\frac{1}{a}\ln |\sin (ax+b)|+C$
• $\int {\sec }^2(ax+b)dx=\frac{1}{a}\tan (ax+b)+C$
• $\int {\mathrm{cosec}}^2(ax+b)dx=-\frac{1}{a}\cot (ax+b)+C$
• $\int \mathrm{cosec}(ax+b)\cdot \cot (ax+b)dx=-\frac{1}{a}\mathrm{cosec}(ax+b)+C$
• $\int \sec (ax+b)\cdot \tan (ax+b)dx=\frac{1}{a}\sec (\mathrm{ax}+\mathrm{b})+\mathrm{C}$
• $\int \sec xdx=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right|+C$
• $\int \mathrm{cosec}xdx=\ln |\mathrm{cosec}x-\cot x|+C=\ln \left|\tan \frac{x}{2}\right|+C$$=-\ln |\mathrm{cosec}x+\cot x|+C$
• $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$
•  $\int \frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}{\sec }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left[x+\sqrt{x^2+a^2}\right]+C$
• $\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left[x+\sqrt{x^2-a^2}\right]+C$ 
• $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C$
• $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C$
• $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
• $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln \left(x+\sqrt{x^2+a^2}\right)+C$
• $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln \left(x+\sqrt{x^2-a^2}\right)+C$
• $\int e^{ax}\cdot \sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\sin \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$
• $\int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\cos \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$

5.0Some definite Formulas are

1. Walli’s Theorem:

(a)  ${\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx=\frac{(n-1)(n-3)\dots ..(1\text{ or }2)}{n(n-2)\dots ..(1\text{ or }2)}K$

$\text{ where }K=\{\begin{array}{ll}\pi /2& \text{ if }n\text{ is even }\\ 1& \text{ if }n\text{ is odd }\end{array}$

(b) ${\int }_0^{\pi /2}{\sin }^{n}x\cdot {\cos }^{m}xdx=\frac{[(n-1)(n-3)(n-5)\dots .1\text{ or }2][(m-1)(m-3)\dots .\dots \text{ or }2]}{(m+n)(m+n-2)(m+n-4)\dots .1\text{ or }2}K$

$\text{ Where }K=\{\begin{array}{ll}\frac{\pi }{2}& \text{ if both }m\text{ and }n\text{ are even }(m,n\in N)\\ 1& \text{ otherwise }\end{array}$

6.0Double and Triple Integrals

Double and Triple Integrals extend the concept of single-variable integration to functions of two or three variables. These integrals are essential for calculating areas, volumes, and other quantities in multi-dimensional spaces.

Double Integrals

Double integrals are used to compute the volume under a surface in three-dimensional space or to find the area of a region in a plane. They are represented as:

${\iint }_{D}f(x,y)dA$

Here, D is the region of integration in the xy-plane, and dA represents a small element of area in this plane. The function f(x, y) is integrated over region D. 

Triple Integrals

Triple integrals are used to calculate the volume of a region in three-dimensional space or to integrate functions of three variables over a specified volume. They are represented as:

${\iiint }_{V}f(x,y,z)dV$

Here, V is the region of integration in three-dimensional space, and dV represents a small volume element. The function f(x, y, z) is integrated over region V.

7.0Integral of Inverse Trigonometric Functions 

The integral of inverse trigonometric functions involves integrating functions that contain the inverse trigonometric functions like arcsin(x), arccos(x), arctan(x), and others. These integrals are essential in various fields of calculus and engineering due to their applications in solving problems involving angles and lengths.

Basic Formulas

Here are some fundamental formulas for the integral of inverse trigonometric functions:

• $\int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$
• $\int {\cos }^{-1}xdx=x{\cos }^{-1}x-\sqrt{1-x^2}+C$
• $\int {\tan }^{-1}xdx=x{\tan }^{-1}x-\frac{1}{2}\ln \left|1+x^2\right|+C$
• $\int {\mathrm{cosec}}^{-1}xdx=x{\mathrm{cosec}}^{-1}x+\ln \left|x+\sqrt{x^2-1}\right|+C$
• $\int {\mathrm{cosec}}^{-1}xdx=x{\mathrm{cosec}}^{-1}x+\ln \left|x+\sqrt{x^2-1}\right|+C$
• $\int {\cot }^{-1}xdx=x^{-0}{\cot }^{-1}x+\frac{1}{2}\ln \left|1+x^2\right|+C$

8.0Solved Examples on Integrals

Example 1:

Evaluate (i) $\int \frac{dx}{8x+3}$ (ii) $\int e^{8x+9}dx$

Solution:

(i) $\frac{1}{8}\ln |8x+9|+C$ (ii)$\frac{e^{8x+9}}{8}+C$

Example 2:

$\int {\cos }^{3}xdx$

Solution:

$\int \frac{3\cos x+\cos 3x}{4}dx\left[\cos 3x=4{\cos }^{3}x-3\cos x\right]$

⇒ $\int \frac{3\cos x}{4}dx+\int \frac{\cos 3x}{4}dx$

⇒$\frac{3\sin x}{4}+\frac{\sin 3x}{12}+C$

Example 3:

$\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Solution:

$\text{ Let }{\sin }^{-1}x=t$

$\int \frac{1}{\sqrt{1-x^2}}dx=dt$

$\Rightarrow \int tdt=\frac{t^2}{2}+C\Rightarrow {\left({\sin }^{-1}x\right)}^2+C$

Example 4:

$\int \frac{9x^2+2}{3x^3+2x+10}dx$

Solution:

$\text{ Put }3x^3+2x+10=t$

$\left(9x^2+2\right)dx=dt$

$=\int \frac{dt}{t}=\ln t+C$

$=\ln \left(3x^2+2x+10\right)+C$

Example 5:

$\int \frac{e^{2x}(1+2x)}{\sin \left(x{e}^{2x}\right)}dx$

Solution:

$\text{ Put, }x{e}^{2x}=t$

$e^{2x}+2x{e}^{2x}dx=dt$

$e^{2x}(1+2x)dx=dt$

⇒ $\int \frac{dt}{\sin t}=\int \mathrm{cosec}tdt$

$\ln \left|\tan \frac{t}{2}\right|+C=\ln \left|\tan \frac{x{e}^{2x}}{2}\right|+C$

Example 6:

Evaluate $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\cdot \frac{1}{x}dx$

Solution:

$\text{ Put }x={\cos }^2\theta \Rightarrow dx=-2\sin \theta \cos \theta d\theta$

$\Rightarrow I=\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\cdot \frac{1}{{\cos }^2\theta }(-2\sin \theta \cos \theta )d\theta =-\int 2\tan \frac{\theta }{2}\tan \theta d\theta$

$\Rightarrow I=\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\cdot \frac{1}{{\cos }^2\theta }(-2\sin \theta \cos \theta )d\theta =-\int 2\tan \frac{\theta }{2}\tan \theta d\theta$

$=-4\int \frac{{\sin }^2(\theta /2)}{\cos \theta }d\theta =-2\int \frac{1-\cos \theta }{\cos \theta }d\theta =-2\ln |\sec \theta +\tan \theta |+2\theta +C$

$=-2\ln \left|\frac{1+\sqrt{1-x}}{\sqrt{x}}\right|+2{\cos }^{-1}\sqrt{x}+C$

Example 7:

${\int }_0^1\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Solution:

${\sin }^{-1}x=t$

$\frac{dx}{\sqrt{1-x^2}}=dt$

at x = 0, t = 0

at x = 1, t=$\pi /2$

${\int }_0^{\pi /2}tdt={\left(\frac{t^2}{2}\right)}_0^{\pi /2}=\frac{{\pi }^2}{8}$

Example 8:

${\int }_0^{\ln 2}\frac{e^x}{1+e^x}dx$

Solution:

$1+e^x=tx=0\Rightarrow t=2$

$e^{x}dx=dtx=\ln 2\Rightarrow t=1+2=3$

${\int }_2^3\frac{dt}{t}\Rightarrow (\ln |t|{)}_2^3=\ln \left(\frac{3}{2}\right)$

Example 9:

${\int }_0^{\pi /2}{\cos }^{6}xdx$

Solution:

n=6, $=\frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot \frac{\pi }{2}=\frac{15\pi }{96}=\frac{5\pi }{32}$

Example 10:

${\int }_0^1\frac{{\tan }^{-1}x}{x}dx=\lambda {\int }_0^{\pi /2}\frac{\theta }{\sin \theta }d\theta \text{, then find }\lambda$

Solution:

L.H.S. $\int \frac{{\tan }^{-1}x}{x}dx$

$\begin{array}{ll}\text{ Put }& {\tan }^{-1}x=t\\ & x=\tan t\\ & dx={\sec }^{2}tdt\end{array}$

$={\int }_0^{\pi /4}\frac{\mathrm{t}}{\tan t}\cdot {\sec }^{2}tdt={\int }_0^{\pi /4}\frac{\mathrm{t}}{\sin t\cos t}dt={\int }_0^{\pi /4}\frac{2\mathrm{t}dt}{\sin 2t}dt$

$\begin{array}{rl}& 2t=\theta \\ & 2dt=d_{\theta }\end{array}$

$=\frac{1}{2}{\int }_0^{\pi /2}\frac{\theta }{\sin \theta }d\theta$

∴ $\lambda =\frac{1}{2}$

Example 11:

Evaluate ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos xdx.$

Solution:

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos xdx=2{\int }_0^{\frac{\pi }{2}}\cos xdx=2(\because \cos x\text{ is even function })$

Example 12:

Evaluate  ${\int }_{-1}^1{\log }_e\left(\frac{2-x}{2+x}\right)dx$. 

Solution:

Let $f(x)={\log }_e\Rightarrow f(-x)={\log }_e\left(\frac{2+x}{2-x}\right)=-{\log }_e\left(\frac{2-x}{2+x}\right)=-f(x)$ 

i.e. f(x) is odd function ∴ ${\int }_{-1}^1{\log }_e\left(\frac{2-x}{2+x}\right)dx=0$

Example 13:

Evaluate ${\int }_0^2\frac{dx}{\left(17+8x-4x^2\right)\left[e^{6(1-x)}+1\right]}$

Solution:

Let I=${\int }_0^2\frac{dx}{\left(17+8x-4x^2\right)\left[e^{6(1-x)}+1\right]}$ 

Also I=${\int }_0^2\frac{dx}{\left(17+8x-4x^2\right)\left[e^{-6(1-x)}+1\right]}$

Adding, we get 

$2\mathrm{I}={\int }_0^2\frac{1}{17+8x-4x^2}\left(\frac{1}{e^{6(1-x)}+1}+\frac{1}{e^{-6(1-x)}+1}\right)dx$

$={\int }_0^2\frac{1}{17+8x-4x^2}dx=-\frac{1}{4}{\int }_0^2\frac{dx}{x^2-2x-17/4}$

$=-\frac{1}{4}{\int }_0^2\frac{dx}{(x-1{)}^2-21/4}=-\frac{1}{4}\times \frac{1}{2\times \frac{\sqrt{21}}{2}}{\left[\log \left|\frac{x-1-\frac{\sqrt{21}}{2}}{x-1+\frac{\sqrt{21}}{2}}\right|\right]}_0^2$

$=-\frac{1}{4\sqrt{21}}{\left[\log \left|\frac{2x-2-\sqrt{21}}{2x-2+\sqrt{21}}\right|\right]}_0^2\Rightarrow |I=$

$=-\frac{1}{4\sqrt{21}}\left[\log \left|\frac{\sqrt{21}-2}{2+\sqrt{21}}\right|\right]$

9.0Practice Problems on Integrals

• $\int \frac{\cos 2x}{{\cos }^{2}x{\sin }^{2}x}dx$
• $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx(\cos x+\sin x>0)$
• $\int \frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$
• $\int \frac{1-\cos x}{1+\cos x}dx$
• $\int {\cos }^{4}xdx$
• ${\int }_a^b\frac{dx}{\sqrt{1+x^2}}where\mathrm{a}=\frac{e-e^{-1}}{2}\&\mathrm{b}=\frac{e^2-e^{-2}}{2}$
• ${\int }_0^{\pi /4}\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }d\theta$
• ${\int }_a^b\frac{|x|}{x}dx$
• ${\int }_0^{\pi /4}(x\cos x\cdot \cos 3x)dx$
• ${\int }_2^3\frac{dx}{(x-1)\sqrt{x^2-2x}}$

10.0Sample Questions on Integrals

Q. How do you evaluate a definite integral?

Ans: To evaluate a definite integral ${\int }_a^{b}f(x)dx$ , find the antiderivative F(x) of f(x) and compute F(b) – F(a).

Q. What is the Fundamental Theorem of Calculus?

Ans: The Fundamental Theorem of Calculus connects differentiation and integration. It states that if F(x) is the antiderivative of f(x), then:

${\int }_a^{b}f(x)dx=F(b)-F(a)$

It also states that the derivative of the integral function is the original function, i.e., if

$F(x)={\int }_a^{x}f(t)dt$ , then F'(x) = f(x).

Q. What are double and triple integrals?

Ans: Double integrals are used to compute the volume under a surface in a two-dimensional region, while triple integrals are used to calculate the volume in a three-dimensional region. They are represented as $\iint$ and $\iiint$ respectively.