Mathematics

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


ANSWER

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


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
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