_0^ xf( x),dx is equal to:

Question

$\displaystyle\int_0^\pi xf(\sin x)\,dx$ is equal to:

Accepted Answer

$$I = \int\limits_0^\pi  {xf\left( {\sin \,x} \right)dx} $$

$$ = \int\limits_0^\pi  {\left( {\pi  - x} \right)f\left( {\sin x} \right)dx} $$

$$ = \pi \int\limits_0^\pi  {f\left( {\sin x} \right)dx - 1} $$

$$ \Rightarrow 2I = \pi {\pi  \over 0}f\left( {\sin x} \right)dx$$

$$I = {\pi  \over 2}\int\limits_0^\pi  {f\left( {\sin x} \right)dx} $$

$$ = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} $$

$$ = \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$

Solutions · 1