Evaluate the following intergrals
(i) `int_(15)^(30) cos (4x - 3)dx` (ii) `int_0^(30) cos 5 x dx` (iii) `int_(0)^(30) cos 5 x dx` (iv) `int_(0)^(10) sec^2 (3x +6)dx`

Answer

(i)`(1)/(4) [sin 117^@] 0.2666` (iii) `(1)/(10)` (iv) `(1)/(3) (tan36^2 - tan 6^@)`

Explanation

(i) Put `4x - 3 =z, so 4 dx =dz or dx = dz//4`
when `x = 15 , z = 4 xx 15 -3 = 57` when `x = 30 , z = 4xx30 -3 +117`
`:. int_(15)^(30) sin 7 x dx = ((-cos 7x)/(7))_0^(30) = (1)/(7)[ cos 7xx30 -cos 7xx10]`
`= - (1)/(7) [cos 210 - cos 0] = -(1)/(7) [cos (180 + 30) - 1]`
`= -(1)/(7)[-cos 30 - 1] = (1)/(7)[sqrt3/(2) +1] = 0.2666`
(iii) `int_0^(30) cos 5 x dx = ((sin 5x)/(5))_0^(30) = (1)/(5)[ sin 5xx30 - sin 5xx10] = (1)/(5)sin 150^@`
`=(1)/(5) sin (180^@- 30^@) = (1)/(5) sin 30^@ = (1)/(5)xx(1)/(2) = (1)/(10)`
(iv) `int_0^(10)sec^2(3x+6)dx =[tan(3x+6)/(3)]_0^(10) =(1)/(3) [ tan (3xx10+6) -tan 6]`
` = (1)/(3) [tan 36^@ - tan 6^@`]`