int_0^{pi / 2} frac{cos x sin x}{1+sin x} d x | vedclass

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(D) Let $I = \int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$. We can rewrite the numerator as $\cos x(1+\sin x - 1)$. Then,$I = \int_0^{\pi / 2}

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$1-\log 2$

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(D) Let $I = \int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$. We can rewrite the numerator as $\cos x(1+\sin x - 1)$. Then,$I = \int_0^{\pi / 2} \frac{\cos x(1+\sin x) - \cos x}{1+\sin x} d x$. This simplifies to $I = \int_0^{\pi / 2} \cos x d x - \int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x$. Evaluating the first integral: $\int_0^{\pi / 2} \cos x d x = [\sin x]_0^{\pi / 2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$. For the second integral,let $t = 1 + \sin x$,then $dt = \cos x dx$. When $x = 0, t = 1$; when $x = \pi/2, t = 2$. So,$\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x = \int_1^2 \frac{dt}{t} = [\log |t|]_1^2 = \log 2 - \log 1 = \log 2$. Therefore,$I = 1 - \log 2$.

$\int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$ is equal to

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