JEE MAIN - Mathematics (2006 - No. 9)
$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to
$$\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx} $$
$$\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx} $$
$${\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx} $$
$$\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$
Обяснение
$$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx} $$
$$ = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1} $$
$$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$
$$I = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$
$$ = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1} $$
$$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$
$$I = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} $$
$$ = \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$