Answer
$$\int^{\pi}_08\sin^4xdx=3\pi$$
Work Step by Step
$$A=\int^{\pi}_08\sin^4xdx$$
This is Case 3, so we need to find a way to rewrite $$(\sin^2x)^k=\Big(\frac{1-\cos2x}{2}\Big)^k$$
We have, $$A=\int^\pi_08(\sin^2x)^2dx$$ $$A=\int^\pi_08\Big(\frac{1-\cos2x}{2}\Big)^2dx$$ $$A=\int^\pi_02(1-\cos2x)^2dx$$ $$A=2\int^\pi_0(1-2\cos2x+\cos^22x)dx$$ $$A=2\Big(x-\sin2x\Big)\Big]^\pi_0+2\int^\pi_0\cos^22xdx$$ $$A=2\Big(\pi-\sin2\pi-0\Big)+2\int^\pi_0\cos^22xdx$$ $$A=2\pi+2\int^\pi_0\cos^22xdx$$
The integral is again an example of Case 3, so we would rewrite $$\cos^22x=\frac{1+\cos4x}{2}$$
$$A=2\pi+2\int^\pi_0\frac{1+\cos4x}{2}dx$$ $$A=2\pi+2\Big(\frac{x}{2}+\frac{1}{8}\sin4x\Big)\Big]^\pi_0$$ $$A=2\pi+2\Big(\frac{\pi}{2}+\frac{\sin4\pi}{8}-0\Big)$$ $$A=2\pi+2\Big(\frac{\pi}{2}\Big)=3\pi$$
