Answer
$$ \frac{1}{4}\left(\frac{3}{4}(2x)+\sin 2x +\frac{1}{4}\sin 2x \cos 2x\right)+c$$
Work Step by Step
\begin{aligned} \int \cos ^{4} x d x &=\int\left(\frac{1}{2}(1+\cos 2 x)\right)^{2} d x \\ &=\frac{1}{4} \int(1+\cos 2 x)^{2} d x \\ &=\frac{1}{4} \int\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x \\
&=\frac{1}{4} \int\left(1+2 \cos 2 x+\frac{1}{2}+ \frac{1}{2}\cos 4 x\right) dx\\
&=\frac{1}{4}\left(\frac{3}{2}x+\sin 2x +\frac{1}{8}\sin 4x \right)+c\\
&= \frac{1}{4}\left(\frac{3}{4}(2x)+\sin 2x +\frac{1}{4}\sin 2x \cos 2x\right)+c
\end{aligned}
