Biết .$\large \int_0^\pi f(\sin x) dx= 1$. Tính $\large \int_0^\pi xf

Biết .$\large \int_0^\pi f(\sin x) dx= 1$. Tính $\large \int_0^\pi xf(\sin x)dx$

• A.

  A. $\large 0$

• B.

  B. $\large \dfrac{1}{2}$

• C.

  C. $\large \dfrac{\pi}{2}$

• D.

  D. $\large \pi$

Chọn C

Đặt $\large x= \pi - t\Rightarrow dx = -dt$. Đổi cận $\large x= 0 \Rightarrow t = \pi;\, x= \pi \Rightarrow  t = 0$

Khi đó: $\large \int_0^\pi xf(\sin x)dx = \int_\pi ^0(\pi - t) f(\sin (\pi -t))(-dt)$

$\large = \int_0^\pi (\pi -t)f(\sin t)dt = \pi \int_0^\pi f(\sin x)dx- \int_0^\pi xf(\sin x)dx$

Do đó: $\large 2\int_0^\pi xf(\sin x)dx= \pi \int_0^\pi f(\sin x)dx= \pi$

Vậy $\large \int_0^\pi xf(\sin x)dx= \dfrac{\pi}{2}$