Mathematics

# Evaluate $\displaystyle \int { \frac { \cos { x } +x\sin { x } }{ { x }^{ 2 }+\cos ^{ 2 }{ x } } dx }$

$-\tan^{-1}\left(\dfrac { \cos { x } }{ x } \right)+c$

##### SOLUTION
$\displaystyle\int \dfrac{\cos x+x\sin x}{x^2+\cos^2x}dx$

$=\displaystyle\int \dfrac{\dfrac{\cos x+x\sin x}{x^2}}{1+\dfrac{\cos^2x}{x^2}}dx$

Put $t=\dfrac{\cos x}{x}$

$\dfrac{dt}{dx}=\dfrac{-x\sin x-\cos x}{x^2}$

$-dt=\dfrac{\cos x+x\sin x}{x^2}dx$

$=\displaystyle\int \dfrac{-dt}{1+t^2}=-\tan^{-1}t+c$

$=-\tan^{-1}\dfrac{\cos x}{x}+C$.

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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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