Tích phân \(\int\limits^{\frac{\pi}{2}}_0\sin^5x\text{d}x\) bằng
\(\dfrac{1}{5}\).\(\dfrac{2}{15}\).\(\dfrac{8}{15}\).\(\dfrac{4}{15}\).Hướng dẫn giải:Đặt \(t=-\cos x\) thì \(\text{d}t=\sin x,\sin^4x=\left(1-t^2\right)^2=t^4-2t^2+1.\) Đổi cận: \(x=|^{\frac{\pi}{2}}_0\Rightarrow t=|^0_1\) suy ra
\(\int\limits^{\frac{\pi}{2}}_0\sin^5x\text{d}x=\int^0_{-1}\left(t^4-2t^2+1\right)\text{d}t=\left(\dfrac{t^5}{5}-\dfrac{2t^3}{3}+t\right)|^0_{-1}=\dfrac{8}{15}.\)