JEE Main · 2006 · 30 Apr 2006 · Offline · Definite Integrals

Apply the reflection property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$ with $a=\pi$. Since $\sin (\pi -x)=\sin x$, the integral $I={\int }_0^{\pi }xf(\sin x)dx$ also equals ${\int }_0^{\pi }(\pi -x)f(\sin x)dx$. Adding the two forms gives $2I=\pi {\int }_0^{\pi }f(\sin x)dx$, so $I=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx$. The graph of $f(\sin x)$ is symmetric about $x=\frac{\pi }{2}$, so ${\int }_0^{\pi }f(\sin x)dx=2{\int }_0^{\pi /2}f(\sin x)dx$ and hence $I=\pi {\int }_0^{\pi /2}f(\sin x)dx$. Finally, over $[0,\frac{\pi }{2}]$ the substitution $x\mapsto \frac{\pi }{2}-x$ replaces $\sin$ by $\cos$, giving the equivalent form $I=\pi {\int }_0^{\pi /2}f(\cos x)dx$.