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(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$. We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$. We

Vedclass · Accepted Answer

$\frac{\sin x-x \cos x}{x \sin x+\cos x}+C$

Vedclass · Answer

(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$. We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$. We can rewrite the integral as: $I = \int \left( \frac{x}{\cos x} ight) \left( \frac{x \cos x}{(x \sin x+\cos x)^{2}} ight) d x$. Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x$. Then $du = \frac{\cos x - x(-\sin x)}{\cos^{2} x} d x = \frac{\cos x + x \sin x}{\cos^{2} x} d x$ and $v = \frac{-1}{x \sin x + \cos x}$. $I = \left( \frac{x}{\cos x} ight) \left( \frac{-1}{x \sin x + \cos x} ight) - \int \left( \frac{\cos x + x \sin x}{\cos^{2} x} ight) \left( \frac{-1}{x \sin x + \cos x} ight) d x$. $I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \frac{1}{\cos^{2} x} d x$. $I = \frac{-x}{\cos x(x \sin x + \cos x)} + 	an x + C$. $I = \frac{-x + 	an x \cos x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x + \sin x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x + x \sin^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x(1 - \sin^{2} x) + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x \cos^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{\cos x(\sin x - x \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{\sin x - x \cos x}{x \sin x + \cos x} + C$.

$\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$ is equal to

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