Explanation:
(B) : Let,
\(I=\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} d x\)
Put, \(\cos \mathrm{x}=\mathrm{t}\)
\(-\sin \mathrm{xdx}=\mathrm{dt}\)
Now, \(x=0, t=\cos 0=1 \backslash\)
And,
\(\mathrm{x}=\frac{\pi}{2}, \mathrm{t}=\cos \frac{\pi}{2}=0\)
\(\mathrm{I}=-\int_{1}^{0} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=-\left[\tan ^{-1} \mathrm{t}\right]_{1}^{0}-\left[\tan ^{-1} 0-\tan ^{-1} 1\right]\)
\(=-\left[0-\frac{\pi}{4}\right]=\frac{\pi}{4}\)