Mathematics

# $\int { x } sinx{ sec }^{ 3 }xdx$ is equal to

$\cfrac { 1 }{ 2 } \left[ { sec }^{ 2 }x-tanx \right] +c$

##### SOLUTION
Given,

$\int \:x\sin \left(x\right)\sec ^3\left(x\right)dx$

$=\int \dfrac{x\sin \left(x\right)}{\cos ^3\left(x\right)}dx$

$=\int \dfrac{x\tan \left(x\right)}{\cos ^2\left(x\right)}dx$

$=\int \:x\tan \left(x\right)\sec ^2\left(x\right)dx$

$\mathrm{Apply\:Integration\:By\:Parts:}\:u=x,\:v'=\sec ^2\left(x\right)\tan \left(x\right)$

$=\dfrac{1}{2}x\tan ^2\left(x\right)-\int \dfrac{1}{2}\tan ^2\left(x\right)dx$

$=\dfrac{1}{2}x\tan ^2\left(x\right)+\dfrac{1}{2}x-\dfrac{1}{2}\tan x$

$=\dfrac{1}{2}x(\tan ^2x+1)-\dfrac{1}{2}\tan x$

$=\dfrac{1}{2}x\sec ^2x-\dfrac{1}{2}\tan x$

$=\dfrac{1}{2}[x\sec ^2x-\tan x]+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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