NIMCET 2018 — Mathematics PYQ

Concept:
- ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.
- If $f(x)=f(2a-x)$, then ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$.
- $\sin (-\theta )=-\sin \theta$.
- $\sin (n\pi +\theta )=(-1{)}^n\sin \theta$.

Calculation:

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

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

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

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

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

Since $f[\sin (\pi -x)]=f(\sin x)$, using ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$, we get:  

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

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