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

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(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi

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$\frac{\pi}{4}-\frac{1}{2} \log 2$

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(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{1+\cos(\pi/2 - x)+\sin(\pi/2 - x)} dx = \int_0^{\pi / 2} \frac{\cos x}{1+\sin x+\cos x} dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{\sin x + \cos x}{1+\sin x+\cos x} dx$. $2I = \int_0^{\pi / 2} \frac{(1+\sin x+\cos x) - 1}{1+\sin x+\cos x} dx = \int_0^{\pi / 2} (1 - \frac{1}{1+\sin x+\cos x}) dx$. $2I = [x]_0^{\pi / 2} - \int_0^{\pi / 2} \frac{1}{1+2\sin(x/2)\cos(x/2)+2\cos^2(x/2)-1} dx$. $2I = \frac{\pi}{2} - \int_0^{\pi / 2} \frac{1}{2\cos^2(x/2)(	an(x/2)+1)} dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi / 2} \frac{\sec^2(x/2)}{	an(x/2)+1} dx$. Let $u = 	an(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$. $2I = \frac{\pi}{2} - \int_0^1 \frac{1}{u+1} du = \frac{\pi}{2} - [\log|u+1|]_0^1 = \frac{\pi}{2} - \log 2$. Therefore,$I = \frac{\pi}{4} - \frac{1}{2} \log 2$.

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

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