If _0^ xf( x),dx = A_0^/2 f( x),dx, then A is:

Question

If $\displaystyle\int_0^\pi xf(\sin x)\,dx = A\int_0^{\pi/2} f(\sin x)\,dx$, then $A$ is:

Accepted Answer

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

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

$$\therefore$$ $$2I = \pi \int\limits_2^\pi  {f\left( {\sin x} \right)} dx$$

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

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

$$ \Rightarrow A = \pi $$

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