## Thomas' Calculus 13th Edition

Given $$\int_{0}^{\pi} 8 \sin ^{4} y \cos ^{2} y d y$$ So, we have \begin{aligned} I&= \int_{0}^{\pi} 8 \sin ^{4} y \cos ^{2} y d y\\ &= \int_{0}^{\pi} 8 (\sin ^{2} )^2y \cos ^{2} y d y\\ &\text{since} \ \ \sin ^{2} y=\frac{1-\cos 2 y}{2},\ \ \cos ^{2}y=\frac{1+\cos 2 y}{2} ,\text{ we get}\\ I&=8 \int_{0}^{\pi}\left(\frac{1-\cos 2 y}{2}\right)^{2}\left(\frac{1+\cos 2 y}{2}\right) d y\\ &= \int_{0}^{\pi}\left( 1-\cos 2 y \right)^{2}\left( 1+\cos 2 y \right) d y\\ &= \int_{0}^{\pi}\left( 1-2\cos 2 y+\cos^2 2 y \right) \left( 1+\cos 2 y \right) d y\\ &= \int_{0}^{\pi}\left( 1-2\cos 2 y+\cos^2 2 y+\cos 2 y-2\cos^2 2 y+\cos^3 2 y \right) d y\\ &= \int_{0}^{\pi}\left( 1-\cos 2 y - \cos^2 2 y+\cos^3 2 y \right) d y\\ &=\int_{0}^{\pi} d y-\int_{0}^{\pi} \cos 2 y d y-\int_{0}^{\pi} \cos ^{2} 2 y d y+\int_{0}^{\pi} \cos ^{3} 2 y d y\\ &=\int_{0}^{\pi} d y-\int_{0}^{\pi} \cos 2 y d y-\int_{0}^{\pi} \cos ^{2} 2 y d y+\int_{0}^{\pi} \cos 2 y\cos ^{2} 2 y d y\\ &=\int_{0}^{\pi} d y-\int_{0}^{\pi} \cos 2 y d y-\int_{0}^{\pi} \cos ^{2} 2 y d y+\int_{0}^{\pi} \cos 2 y(1-\sin ^{2} 2 y) d y\\ &=\int_{0}^{\pi} d y -\int_{0}^{\pi} \cos ^{2} 2 y d y-\int_{0}^{\pi} \cos 2 y \sin ^{2} 2 y d y\\ &\text{since} \ \ \cos ^{2} 2 y=\frac{1+\cos 4 y}{2} ,\text{ we get}\\ I&=\left[y \right]_{0}^{\pi}-\int_{0}^{\pi}\left(\frac{1+\cos 4 y}{2}\right) d y-\int_{0}^{\pi} \cos 2 y \sin ^{2} 2 y d y\\ &=\pi-\frac{1}{2} \int_{0}^{\pi} d y-\frac{1}{2} \int_{0}^{\pi} \cos 4 y d -\int_{0}^{\pi} \cos 2 y \sin ^{2} 2 y d y\\ &=\pi+\left[-\frac{1}{2} y-\frac{1}{8} \sin 4 y -\frac{1}{2} \cdot \frac{\sin ^{2} y}{3}\right]_{0}^{\pi}\\ &=\pi+\left[-\frac{1}{2} \pi-\frac{1}{8} \sin 4\pi -\frac{1}{2} \cdot \frac{\sin ^{2} \pi}{3}\right]- \left[0-\frac{1}{8} \sin 0 -\frac{1}{2} \cdot \frac{\sin ^{2} 0}{3}\right] \\ &=\pi-\frac{\pi}{2}\\ &=\frac{\pi}{2} \\ \end{aligned}