Tinh tích phân 0 đến pi x(x+sinx)dx

$\begin{array}{l} I = \int\limits_0^\pi {{x^2}dx} + \int\limits_0^\pi {x\sin xdx} = \left. {\frac{1}{3}{x^3}} \right|_0^\pi + \int\limits_0^\pi {x\sin xdx} \\ = \frac{1}{3}{\pi ^3} + \int\limits_0^\pi {x\sin xdx} \end{array}$

Tính $\int\limits_0^\pi {x\sin xdx}$  

Đặt: $\left\{ \begin{array}{l} u = x\\ dv = \sin xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = - \cos x \end{array} \right.$  

$\begin{array}{l} \int\limits_0^\pi {x\sin xdx} = \left. { - x{\mathop{\rm cosx}\nolimits} } \right|_0^\pi + \int\limits_0^\pi {\cos dx} \\ = \pi + \left. {\sin x} \right|_0^\pi = \pi \end{array}$  

Vậy $I = \frac{1}{3}{\pi ^3} + \pi .$