int frac{x(x sin x+cos x)^{-2}}{sec x} d x= | vedclass

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Question

(B) Let $I = \int \frac{x \cos x}{(x \sin x + \cos x)^2} dx$. Consider $u = x \sin x + \cos x$. Then $du = (1 \cdot \sin x + x \cos x - \sin x) dx = x

Vedclass · Accepted Answer

$\frac{-1}{x \sin x+\cos x}$

Vedclass · Answer

(B) Let $I = \int \frac{x \cos x}{(x \sin x + \cos x)^2} dx$. Consider $u = x \sin x + \cos x$. Then $du = (1 \cdot \sin x + x \cos x - \sin x) dx = x \cos x dx$. Substituting these into the integral,we get: $I = \int \frac{1}{u^2} du = \int u^{-2} du$. Integrating,we get $I = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$. Substituting back $u = x \sin x + \cos x$,we get $I = -\frac{1}{x \sin x + \cos x} + C$. Thus,the correct option is $B$.

$\int \frac{x(x \sin x+\cos x)^{-2}}{\sec x} d x=$ . . . . . . $+C$

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