Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_0^{\pi }xf(\sin x)dx$.

Using the property ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, we have:

$$I={\int }_0^{\pi }(\pi -x)f(\sin (\pi -x))dx$$

$$I={\int }_0^{\pi }(\pi -x)f(\sin x)dx$$

$$I=\pi {\int }_0^{\pi }f(\sin x)dx-{\int }_0^{\pi }xf(\sin x)dx$$

$$I=\pi {\int }_0^{\pi }f(\sin x)dx-I$$

$$2I=\pi {\int }_0^{\pi }f(\sin x)dx$$

$$I=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx$$

Again, using the property ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ if $f(2a-x)=f(x)$, we can write:

$$I=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx=2\frac{\pi }{2}{\int }_0^{\pi /2}f(\sin x)dx$$

$$I=\pi {\int }_0^{\pi /2}f(\sin x)dx$$

Hence, option (C) is the correct answer.