int frac{x^2}{(xsin x + cos x)^2} , dx = | vedclass

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(C) Let $I = \int \frac{x^2}{(x\sin x + \cos x)^2} \, dx$. We can rewrite the integrand as: $I = \int \frac{x \cos x}{(x\sin x + \cos x)^2} \cdot \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{x^2}{(x\sin x + \cos x)^2} \, dx$. We can rewrite the integrand as: $I = \int \frac{x \cos x}{(x\sin x + \cos x)^2} \cdot \frac{x}{\cos x} \, dx$. Let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x\sin x + \cos x)^2} \, dx$. Then $du = \frac{\cos x - x(-\sin x)}{\cos^2 x} \, dx = \frac{\cos x + x\sin x}{\cos^2 x} \, dx$. Integrating $dv$,we get $v = \int \frac{x \cos x}{(x\sin x + \cos x)^2} \, dx$. Let $t = x\sin x + \cos x$,then $dt = (\sin x + x\cos x - \sin x) \, dx = x\cos x \, dx$. So,$v = \int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{x\sin x + \cos x}$. Using integration by parts,$I = uv - \int v \, du$: $I = -\frac{x}{\cos x(x\sin x + \cos x)} - \int \left( -\frac{1}{x\sin x + \cos x} ight) \cdot \frac{x\sin x + \cos x}{\cos^2 x} \, dx$. $I = -\frac{x}{\cos x(x\sin x + \cos x)} + \int \sec^2 x \, dx$. $I = -\frac{x}{\cos x(x\sin x + \cos x)} + 	an x$. $I = \frac{-x + 	an x \cos x (x\sin x + \cos x)}{\cos x (x\sin x + \cos x)}$. $I = \frac{-x + \sin x (x\sin x + \cos x)}{\cos x (x\sin x + \cos x)} = \frac{-x + x\sin^2 x + \sin x \cos x}{\cos x (x\sin x + \cos x)}$. $I = \frac{\sin x \cos x - x(1 - \sin^2 x)}{\cos x (x\sin x + \cos x)} = \frac{\sin x \cos x - x \cos^2 x}{\cos x (x\sin x + \cos x)}$. $I = \frac{\cos x (\sin x - x \cos x)}{\cos x (x\sin x + \cos x)} = \frac{\sin x - x \cos x}{x\sin x + \cos x} + C$.

$\int \frac{x^2}{(x\sin x + \cos x)^2} \, dx = $

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