Đặt $\begin{cases}u=x \\ dv=cosx.dx \end{cases}\Rightarrow \begin{cases}du=dx \\ v=sinx \end{cases}$Suy ra $I=xsinx|_0^\pi-\int\limits_{0}^{\pi}sinxdx=\int\limits_{0}^{\pi}dcosx=\frac{cos^2x}{2}|_0^\pi=0$
Đặt $\begin{cases}u=x \\ dv=cosx.dx \end{cases}\Rightarrow \begin{cases}du=dx \\ v=sinx \end{cases}$Suy ra $I=xsinx|_0^\pi-\int\limits_{0}^{\pi}sinxdx=\int\limits_{0}^{\pi}dcosx=\frac{cosx^2}{2}|_0^\pi=0$
Đặt $\begin{cases}u=x \\ dv=cosx.dx \end{cases}\Rightarrow \begin{cases}du=dx \\ v=sinx \end{cases}$Suy ra $I=xsinx|_0^\pi-\int\limits_{0}^{\pi}sinxdx=\int\limits_{0}^{\pi}dcosx=\frac{cos^2
x}{2}|_0^\pi=0$