Answer to Question #150989 in Calculus for Ambkeshwar

Question

Integrate the following functions w.r.t. x using substitution: 

 i) x x 1dx

1

3 4

∫

+ ii) ∫

π

4

7 2

tan x sec x dx

 iii) ∫

sec (cos )x sin x dx

Answer 1

1) Given $I=\intop _0^1x^3.(x^4+1)dx$

Let, $z=x^4+1$

Then,$dz/dx =4x^3$

$\implies dz/4=x^3dx$

Now,$\intop x^3.(x^4+1)dx=\intop z.(dz/4)$

$=(1/4).(z^2/2)+C$

$=(1/8).(x^4+1)^2+C$

Therefore, $I=\intop _0^1x^3.(x^4+1)dx$

$=[(1/8).(x^4+1)^2+C]_0^1$

$=(1/8).(1^4+1)^2-(1/8).(0^4+1)^2$

$=4.(1/8)-(1/8)$

$=3/8$

2)Given $I=\intop _{0}^{\pi /4} tan^7x.sec^2x dx$

Let, $z=tanx$

Then,${dz}/{dx}=sec^2x$

$\implies dz=sec^2xdx$

Again,when $x\to \pi /4,z\to1$ and

when $x\to 0,z\to 0$

Therefore, $I=\intop _0^{\pi/4}tan^7x.sec^2xdx$

$=\intop _0^1z^7dz$

$=[z^8/8]_0^1$

$=(1/8).[1^8-0]$

$=(1/8)$

3) Given $I=\intop sec(cosx).sinx$ $dx$

Let, $z=cosx$

Then, $dz/dx=(-sinx)$

$\implies (-dz)=sinx$ $dx$

Therefore, $I=\intop sec(cosx).sinx$ $dx$

$=\intop secz (-dz)$

$=-\intop secz$ $dz$

$=-log|secz+tanz|+C$

$=-log|sec(cosx)+tan(cosx)|+C$