int_0^{pi / 4} frac{x^2}{(x sin x+cos x)^2} d | vedclass

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(B) Let $I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx$.We can rewrite the integrand as $I = \int (x \sec x) \left( \frac{x \cos x}{(x \sin x + \cos x

Vedclass · Accepted Answer

$\frac{4-\pi}{4+\pi}$

Vedclass · Answer

(B) Let $I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx$.We can rewrite the integrand as $I = \int (x \sec x) \left( \frac{x \cos x}{(x \sin x + \cos x)^2} ight) dx$.Using integration by parts,let $u = x \sec x$ and $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} dx$.Then $du = (\sec x + x \sec x 	an x) dx$ and $v = \frac{-1}{x \sin x + \cos x}$.$I = (x \sec x) \left( \frac{-1}{x \sin x + \cos x} ight) - \int (\sec x + x \sec x 	an x) \left( \frac{-1}{x \sin x + \cos x} ight) dx$.$I = \frac{-x \sec x}{x \sin x + \cos x} + \int \frac{\sec x (1 + x 	an x)}{x \sin x + \cos x} dx$.Since $x \sin x + \cos x = \cos x (x 	an x + 1)$,the integral becomes $\int \frac{\sec x \cdot \cos x (x 	an x + 1)}{\cos x (x 	an x + 1)} dx = \int \sec^2 x dx = 	an x$.Thus,$I = \frac{-x \sec x}{x \sin x + \cos x} + 	an x = \frac{-x + 	an x (x \sin x + \cos x)}{\cos x (x \sin x + \cos x)} = \frac{\sin x - x \cos x}{x \sin x + \cos x}$.Evaluating the definite integral: $\left[ \frac{\sin x - x \cos x}{x \sin x + \cos x} ight]_0^{\pi / 4} = \frac{\frac{1}{\sqrt{2}} - \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} - 0 = \frac{1 - \pi/4}{1 + \pi/4} = \frac{4 - \pi}{4 + \pi}$.

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

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