Bài 106346

Question

Tính tích phân:

$\int\limits_0^\pi {xsin xcos ^2xdx}$

Tiểu Bắc · Answer 1 · 5/31/2012 1:43:23 PM

I =

$\int\limits_0^\pi {x\sin x{{\cos }^2}xdx}$
Đặt

$x = \pi - t$

$\begin{array}{l} I = - \int\limits_\pi ^0 {(\pi - t)\sin t{{\cos }^2}tdt} = \pi \int\limits_0^\pi {\sin x{{\cos }^2}xdx} - \int\limits_0^\pi {x\sin x{{\cos }^2}xdx} \\ \end{array}$

$I_1=\pi\int^{\pi}_0\sin x\cos^2xdx=-\pi\int^{\pi}_0\cos^2xd\cos x$

$=\frac{-\pi}{3}\cos^3x|^{\pi}_0=\frac{2\pi}{3}$

$I_2=\int^{\pi}_0 x\sin x\cos^2xdx=-\int^{\pi}_0 x\cos^2xd\cos x$

$=-\frac{1}{3}\int^{\pi}_0 xd\cos^3 x=\frac{-1}{3} x\cos^3 x|^{\pi}_0+\frac{1}{3}\int^{\pi}_0 \cos^3 xdx$

$=\frac{\pi}{3}+\frac{1}{12}\cos^4x|^{\pi}_0=\frac{\pi}{3}$

$\Rightarrow I=\frac{\pi}{3}$

Bài 106346

1 Đáp án

Lý thuyết liên quan

Bài 106346

1 Đáp án

Liên quan

Lý thuyết liên quan