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

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

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$\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx$ ...$(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos x + \sin x} dx$ ...(ii)Adding $(i)$ and (ii):$2I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} dx$$2I = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x} dx = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \operatorname{cosec}\left(x + \frac{\pi}{4}ight) dx$$2I = \frac{1}{\sqrt{2}} \left[ \log \left| 	an \left( \frac{x}{2} + \frac{\pi}{8} ight) ight| ight]_0^{\frac{\pi}{2}}$Alternatively,using $\int \operatorname{cosec} 	heta d	heta = \log |\operatorname{cosec} 	heta - \cot 	heta |$:$2I = \frac{1}{\sqrt{2}} \left[ \log \left| \operatorname{cosec} \left( x + \frac{\pi}{4} ight) - \cot \left( x + \frac{\pi}{4} ight) ight| ight]_0^{\frac{\pi}{2}}$$2I = \frac{1}{\sqrt{2}} [ \log |\operatorname{cosec} \frac{3\pi}{4} - \cot \frac{3\pi}{4}| - \log |\operatorname{cosec} \frac{\pi}{4} - \cot \frac{\pi}{4}| ]$$2I = \frac{1}{\sqrt{2}} [ \log |\sqrt{2} - (-1)| - \log |\sqrt{2} - 1| ] = \frac{1}{\sqrt{2}} [ \log (\sqrt{2}+1) - \log (\sqrt{2}-1) ]$Since $\log (\sqrt{2}-1) = \log (\frac{1}{\sqrt{2}+1}) = -\log (\sqrt{2}+1)$:$2I = \frac{1}{\sqrt{2}} [ 2 \log (\sqrt{2}+1) ] \implies I = \frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$.

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx =$

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