Question

Tính tích phân :

\(\int^{\frac{\pi}{2}}_0x^2\cos xdx\)

 

 

Answer 1

Đặt \(u=x^2\rightarrow du=2xdx,dv=\cos xdx\rightarrow v=\sin x\)

Do đó : 

\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)

\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)