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

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Question

(C) We are given the integral $I = \int \frac{x^2 \, dx}{(x \sin x + \cos x)^2}$.We can rewrite the integrand as $I = \int \frac{x}{(x \sin x + \cos x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We are given the integral $I = \int \frac{x^2 \, dx}{(x \sin x + \cos x)^2}$.We can rewrite the integrand as $I = \int \frac{x}{(x \sin x + \cos x)^2} \cdot \frac{x}{\cos x} \, dx$.Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \, dx}{(x \sin x + \cos x)^2}$.Then $du = \frac{\cos x - x(-\sin x)}{\cos^2 x} \, dx = \frac{\cos x + x \sin x}{\cos^2 x} \, dx$.To find $v$,let $t = x \sin x + \cos x$,then $dt = (\sin x + x \cos x - \sin x) \, dx = x \cos x \, dx$. This does not simplify directly,so we use the derivative of $\frac{-1}{x \sin x + \cos x}$ which is $\frac{-(x \cos x)}{(x \sin x + \cos x)^2}$.Thus,$v = \frac{-1}{x \sin x + \cos x} \cdot \frac{1}{\cos x}$ is not quite right. Let's use $I = \int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx$.Let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx$. Then $v = \frac{-1}{x \sin x + \cos x}$.$I = \frac{-x}{\cos x(x \sin x + \cos x)} - \int \left( \frac{-1}{x \sin x + \cos x} ight) \left( \frac{\cos x + x \sin x}{\cos^2 x} ight) \, dx$.$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \sec^2 x \, dx$.$I = \frac{-x}{\cos x(x \sin x + \cos x)} + 	an x + c$.Comparing this with $f(x) + 	an x + c$,we get $f(x) = \frac{-x}{\cos x(x \sin x + \cos x)}$.

If $I = \int \frac{x^2 \, dx}{(x \sin x + \cos x)^2} = f(x) + \tan x + c$,then $f(x)$ is

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