Ta có \(I=\int_{0}^{\frac{\pi}{2}}xsinxdx+\int_{0}^{\frac{\pi}{2}}sinx.cos^5xdx\) Đặt \(\left\{\begin{matrix} u=x\\ du=sinxdx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=-cosx \end{matrix}\right.\) Khi đó \(\int_{0}^{\frac{\pi}{2}}xsinxdx=(-xcosx) \bigg|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}}cosxdx=0+(sinx)\bigg|_{0}^{\frac{\pi}{2}} =1\) Đặt \(t=cosx\Rightarrow dt=-sinxdx\) Đổi cận \(\left\{\begin{matrix} x=\frac{\pi}{2}\\ x=0 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} t=0\\ t=1 \end{matrix}\right.\) Khi đó \(\int_{0}^{\frac{\pi}{2}}sinx.cos^5xdx=\int_{1}^{0}-t^5dt=\int_{0}^{1}t^5dt= \left ( \frac{t^6}{6} \right )\bigg |^1_0=\frac{1}{6}\) Vậy \(I=1+\frac{1}{6}=\frac{7}{6}\)