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.