Answer
$$ \int \cos ^{4} x d x =\frac{1}{4}\cos ^{3} x \sin x +\frac{3}{4} \cos x \sin x+\frac{3}{8} x +C$$
Work Step by Step
$$
\int \cos ^{4} x d x
$$
Here $n=4$ , Use the formula $$
\int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x
$$
We get
\begin{align*}
\int \cos ^{4} x d x&=\frac{\cos ^{3} x \sin x}{4}+\frac{3}{4} \int \cos ^{2} x d x\\
&=\frac{\cos ^{3} x \sin x}{4}+\frac{3}{4}\left(\cos x \sin x+\frac{1}{2} x\right)\\
&=\frac{1}{4}\cos ^{3} x \sin x +\frac{3}{4} \cos x \sin x+\frac{3}{8} x +C
\end{align*}
