Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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Evaluate $\int \dfrac {1}{a^{x}b^{x}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate: $\displaystyle \int \dfrac { 1 } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } } d x ; x = 0$
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