## Thomas' Calculus 13th Edition

Given $$\int_{0}^{\pi} \sin ^{5}\left(\frac{x}{2}\right) \mathrm{d} x$$ So, we have \begin{aligned} I&= \int_{0}^{\pi} \sin ^{5}\left(\frac{x}{2}\right) \ dx \\ &=\int\left(\sin ^{2} \left(\frac{x}{2}\right)\right)^{2} \sin x \ d x\\ &=\int\left(1-\cos ^{2} \left(\frac{x}{2}\right)\right)^{2} \sin x\ d x\\ &=\int\left(1-2 \cos ^{2} \left(\frac{x}{2}\right)+\cos ^{4}\left(\frac{x}{2}\right)\right) \sin x \ d x\\ &=\int_{0}^{\pi} \sin \left(\frac{x}{2}\right) d x-\int_{0}^{\pi} 2 \cos ^{2}\left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) d x+\int_{0}^{\pi} \cos ^{4}\left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) d x\\ &=2\int_{0}^{\pi} \sin \left(\frac{x}{2}\right) \frac{1}{2}d x-4\int_{0}^{\pi} \cos ^{2}\left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) \frac{1}{2}d x+2\int_{0}^{\pi} \cos ^{4}\left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) \frac{1}{2}d x\\ &=\left[-2 \cos \left(\frac{x}{2}\right)+\frac{4}{3} \cos ^{3}\left(\frac{x}{2}\right)-\frac{2}{5} \cos ^{5}\left(\frac{x}{2}\right)\right]_{0}^{\pi}\\ &=\left[-2 \cos \left(\frac{\pi}{2}\right)+\frac{4}{3} \cos ^{3}\left(\frac{\pi}{2}\right)-\frac{2}{5} \cos ^{5}\left(\frac{\pi}{2}\right)\right] - \left[-2 \cos \left(0\right)+\frac{4}{3} \cos ^{3}\left(0\right)-\frac{2}{5} \cos ^{5}\left(0\right)\right]0\\ &=(0)-\left(-2+\frac{4}{3}-\frac{2}{5}\right)\\ &=\frac{16}{15}\\ \end{aligned}