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

Question

Aicharya · Accepted Answer

Let $I=\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x$ ...(i) On applying property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$, we get, $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos x+\sin x} d x$ ...(ii) On adding Eqs. (i) and (ii), we get $\begin{aligned} 2 I & =\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x+\cos ^2 x}{\sin x+\cos x} d x \ & =\int_0^{\frac{\pi}{2}} \frac{d x}{\sin x+\cos x} \ & =\frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \operatorname{cosec}\left(\frac{\pi}{4}+x\right) d x \ & =\frac{1}{\sqrt{2}}\left[\log \left|\operatorname{cosec}\left(\frac{\pi}{4}+x\right)-\cot \left(\frac{\pi}{4}+x\right)\right|\right]_0^{\pi / 2} \ \Rightarrow I & =\frac{1}{2 \sqrt{2}}[\log (\sqrt{2}+1)-\log (\sqrt{2}-1)] \ & =\frac{1}{2 \sqrt{2}}[2 \log (\sqrt{2}+1)]=\frac{1}{\sqrt{2}} \log (\sqrt{2}+1) \end{aligned}$ Hence, option (b) is correct.

\(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x=\)

Solution