intlimits_0^pi {frac{{xcos x}}{{{{left( {1 + | vedclass

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(C) Let $I = \int\limits_0^\pi  {\frac{{x\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx

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$0$

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(C) Let $I = \int\limits_0^\pi  {\frac{{x\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi \frac{(\pi - x)\cos(\pi - x)}{(1 + \sin(\pi - x))^2} dx = \int_0^\pi \frac{(\pi - x)(-\cos x)}{(1 + \sin x)^2} dx$. $I = -\pi \int_0^\pi \frac{\cos x}{(1 + \sin x)^2} dx + \int_0^\pi \frac{x\cos x}{(1 + \sin x)^2} dx$. $I = -\pi \int_0^\pi \frac{\cos x}{(1 + \sin x)^2} dx + I$. This implies $\pi \int_0^\pi \frac{\cos x}{(1 + \sin x)^2} dx = 0$. Let $u = 1 + \sin x$,then $du = \cos x dx$. When $x = 0, u = 1$. When $x = \pi, u = 1$. Thus,$\int_1^1 \frac{1}{u^2} du = 0$. Therefore,the value of the integral is $0$.

$\int\limits_0^\pi {\frac{{x\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}} dx$ is equal to :

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