Mathematics

$$\int x\sin x\sec^{3}xdx=$$


ANSWER

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


SOLUTION

Consider the following integral .

  $$ I=\int xsinxse{{c}^{3}}xdx=\int\limits_{{}}^{{}}{x\tan x{{\sec }^{2}}x} $$

let $$ t=\tan x $$ and differentiate both side w.r.t x, we get.

 $$ \dfrac{dt}{{{\sec }^{2}}x}=dx $$

 $$ =\int\limits_{{}}^{{}}{\dfrac{t{{\tan }^{-1}}t{{\sec }^{2}}x}{{{\sec }^{2}}x}}dt $$

 $$ =\int\limits_{{}}^{{}}{t{{\tan }^{-1}}t}dt $$

 $$ ={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\int\limits_{{}}^{{}}{\dfrac{1}{1+{{t}^{2}}}}}{{t}^{2}}dt $$

 $$ ={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\left( \int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt} \right)} $$

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

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

Hence, this is the correct answer.

.   

 

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Single Correct Medium Published on 17th 09, 2020
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