Evaluate x^2 d x (x x+ x)^2 . - Vidyadip

Question

Evaluate $\int \frac{x^2 d x}{(x \sin x+\cos x)^2}$.

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Answer

Let
$I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}$
Since $\frac{d}{d x}(x \sin x+\cos x)=1 \cdot \sin x+x \cos x-\sin x$
$=x \cos x$
So assuming $x^2$ as $\frac{x}{\cos x} \cdot x \cos x$ taking
$\frac{x \cos x}{(x \sin x+\cos x)^2}$ as the second function, integrating by parts,
$\int \frac{x^2}{(x \sin x+\cos x)^2} d x$
$=\int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x$
$=\frac{x}{\cos x} \times\left(\frac{-1}{x \sin x+\cos x}\right)-$
$\int \frac{\cos x+x \sin x}{\cos ^2 x} \times\left\{\frac{-1}{x \sin x+\cos x}\right\} d x$
$=\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^2 x d x$
$=\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+ C$
$=\frac{-x}{\cos x(x \sin x+\cos x)}+\frac{\sin x}{\cos x}+ C$
$=\frac{-x+x \sin ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C $
$=\frac{-x \cos ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C$
$=\frac{\sin x-x \cos x}{x \sin x+\cos x}+ C \text { }$

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