Question

Tính tích phân :

                       \(I=\int\limits^{\pi}_0x\left(x-\sin x\right)dx\)

Answer 1

\(I=\int\limits^{\pi}_0\left(x^2-x\sin x\right)dx=\frac{x^3}{3}|^{\pi}_0-\int^{\pi}_0x\sin xdx=\frac{\pi^3}{3}-\int\limits^{\pi}_0x\sin xdx\)

Tính \(I_1=\int\limits^{\pi}_0x\sin xdx\)

Đặt \(\begin{cases}u=x\\dv=\sin xdx\end{cases}\)\(\Rightarrow\begin{cases}du=dx\\v=-\cos x\end{cases}\)

\(\Rightarrow I_1=-x\cos x|^{\pi}_0+\int\limits^{\pi}_0\cos xdx=\pi+\sin x|^{\pi}_0=\pi\)

\(\Rightarrow I=\frac{\pi^3}{3}-\pi\)