Explanation:
(B) :
\(I=\int_{0}^{\pi} f(\sin x) d x\)
We know that,
\(\int_{0}^{a} f(x) d x=\int_{0}^{a}(f(x)+f(2 a-x)) d x\)
So,
\(I=\int_{0}^{\pi / 2}(f(\sin x)+f(\sin (\pi-x))) d x\)
\(I=\int_{0}^{\pi / 2}(f(\sin x)+f(\sin x)) d x=2 \int_{0}^{\pi / 2} f(\sin x) d x\)
or \(\int_{0}^{\pi} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}=2 \int_{0}^{\pi / 2} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}\)
or \(\frac{\int_{0}^{\pi} f(\sin x) d x}{\int_{0}^{\pi / 2} f(\sin x) d x}=2\)
Hence, \(\int_{0}^{\pi} f(\sin x) d x: \int_{0}^{\pi / 2} f(\sin x) d x=2\)