NIMCET 2018 Question Paper with solutions

Given,
I = $\underset{0}{\overset{\mathit{\pi }}{\int }}$ x f (sin x) dx
= $\underset{0}{\overset{\mathit{\pi }}{\int }}$ (π - x) f {sin (π - x)} dx
= $\underset{0}{\overset{\mathit{\pi }}{\int }}$ (π - x) f (sin x) dx
= $\underset{0}{\overset{\mathit{\pi }}{\int }}$ π f (sin x) dx - $\underset{0}{\overset{\mathit{\pi }}{\int }}$ x f (sin x) dx
= π $\underset{0}{\overset{\mathit{\pi }}{\int }}$ f (sin x) dx - I
⇒ 2I = π $\underset{0}{\overset{\mathit{\pi }}{\int }}$ [f (sin x) + f (sin (π - x)] dx
= 2π $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ f sin $\left(\frac{\mathit{\pi }}{2}-\mathit{x}\right)$ dx
⇒ I = π $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ f (cos) x dx