JEE MAIN - Mathematics (2004 - No. 20)
If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } $$ then $$A$$ is
$$2\pi $$
$$\pi $$
$${\pi \over 4}$$
$$0$$
ব্যাখ্যা
Let $$I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$$
$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$
$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$
$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
$$ \Rightarrow A = \pi $$
$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$
$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$
$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
$$ \Rightarrow A = \pi $$