Explanation:
(C) : Given,
\(I=\int_{0}^{\pi / 2} \frac{\sin 2 x}{1+2 \cos ^{2} x} d x\)
\(=\int_{0}^{\pi / 2} \frac{\sin 2 x}{1+2 \cos ^{2} x-1+1} d x\)
\(=\int_{0}^{\pi / 2} \frac{\sin 2 x}{2+\cos 2 x} d x\)
\(\left(2 \cos ^{2} x-1=\cos 2 x\right)\)
\(2+\cos 2 x=t\)
\(-2 \sin 2 x d x=d t\)
\(\sin 2 x d x=\frac{-1}{2} \quad d t\)
When, \(\mathrm{x}=0, \mathrm{t}=3\)
\(x=\frac{\pi}{2}, t=1\)
Now, \(I=\frac{-1}{2} \int_{3}^{1} \frac{1}{t} \mathrm{dt}\)
\(=\frac{-1}{2}[\log \mathrm{t}]_{3}^{1}\)
\(=\frac{-1}{2}[\log 1-\log 3]=\frac{1}{2} \log 3\)