The value of ∫_0^(π)/2sin^5x dx is

The correct answer is Option (4) → \(\displaystyle\frac{8}{15}\)

Since the power of \(\sin x\) is odd, keep one \(\sin x\) for dt and write the rest in \(\cos x\):

    \( \sin^{5}x = \left(1-\cos^{2}x\right)^{2}\sin x \)

Let \(t = \cos x\). Then \(dt = -\sin x\,dx\).

When \(x = 0\), \(t = 1\); when \(x = \frac{\pi}{2}\), \(t = 0\). So

    \( \displaystyle \int_{0}^{\pi/2}\sin^{5}x\,dx = -\int_{1}^{0}\left(1-t^{2}\right)^{2}dt \)

    \( \displaystyle = \int_{0}^{1}\left(1 - 2t^{2} + t^{4}\right)dt \)

    \( \displaystyle = 1 - \frac{2}{3} + \frac{1}{5} \)

    \( \displaystyle = \frac{15 - 10 + 3}{15} = \frac{8}{15} \)