Explanation:
(A) : Let,
\(I=\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x\)
Put, \(\sin ^{2} \mathrm{x}=\mathrm{t}\)
\(2 \sin x \cos x=d t\)
Here, if \(\mathrm{x}=0\) we get \(\mathrm{t}=0\)
And, if \(x=\pi / 2\) we get \(t=1\)
Substituting the values,
\(\mathrm{I}=\frac{1}{2} \int_{0}^{1} \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}\)
Integrating, w.r.t. \(t\) we get -
\(=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1}\)
\(=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]=\frac{1}{2} \times \frac{\pi}{4}\)
So, \(\quad \mathrm{I}=\frac{\pi}{8}\)