Given int_{0}^{frac{pi}{2}} frac{dx}{1 + sin | vedclass

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Question

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x + \cos x} dx$ and $J = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x + \cos x} dx$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x + \cos x} dx$ and $J = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x + \cos x} dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get $J = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos x + \sin x} dx = I$.Now,$I + J = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{1 + \sin x + \cos x} dx$.Since $I = J$,we have $2I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \sin x + \cos x - 1}{1 + \sin x + \cos x} dx = \int_{0}^{\frac{\pi}{2}} (1 - \frac{1}{1 + \sin x + \cos x}) dx$.$2I = [x]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin x + \cos x}$.$2I = \frac{\pi}{2} - \ln 2$.$I = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

Given $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin x + \cos x} = \ln 2$,then the value of the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x + \cos x} dx$ is equal to

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