Answer
\begin{aligned}
\int_{0}^{\pi} 8 \sin ^{4} x d x =3 \pi
\end{aligned}
Work Step by Step
Given $$ \int_{0}^{\pi} 8 \sin ^{4} x d x $$
So, we have
\begin{aligned}
I&= \int_{0}^{\pi} 8 \sin ^{4} x d x\\
&= \int_{0}^{\pi} 8 (\sin ^{2} x)^2 d x, \\
&\text{since} \ \ \sin ^{2} x=\frac{1-\cos 2 x}{2} ,\text{ we get}\\
I&=8 \int_{0}^{\pi}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x\\
&=2 \int_{0}^{\pi}\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x\\
&\text{since} \ \ \cos ^{2} 2x=\frac{1+\cos 4 x}{2} ,\text{ we get}\\
I&=2 \int_{0}^{\pi} d x-2 \int_{0}^{\pi} \cos 2 x \cdot 2 d x+2 \int_{0}^{\pi} \frac{1+\cos 4 x}{2} d x\\
&=[2 x-2 \sin 2 x]_{0}^{\pi}+\int_{0}^{\pi} d x+\int_{0}^{\pi} \cos 4 x d x\\
&=2 \pi-2\sin 2\pi-0+2\sin0+\left[x+\frac{1}{2} \sin 4 x\right]_{0}^{\pi}\\
&=2 \pi+\left[\pi+\frac{1}{2} \sin 4 \pi\right]-\left[0+\frac{1}{2} \sin 0\right]\\
&=2 \pi+\pi\\
&=3 \pi
\end{aligned}
