Answer
$$\frac{1}{12} \cos ^{3}(3 x+2) \sin (3 x+2)+\frac{1}{8}(3 x+2)+\frac{1}{16} \sin (6 x+4)+c$$
Work Step by Step
Given
$$\int \cos ^{4}(3 x+2) d x $$
Let
$$ u=3x+2 \ \ \ \Rightarrow \ \ \ du=3dx $$
Then
$$\int \cos ^{4}(3 x+2) d x=\frac{1}{3} \int \cos ^{4} u d u$$
Hence by using
$$ \int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$$
We get
\begin{align*}
\int \cos ^{4}(3 x+2) d x&=\frac{1}{3} \int \cos ^{4} u d u\\
&= \frac{1}{3}\left(\frac{1}{4} \cos ^{3} u \sin u+\frac{3}{4} \int \cos ^{2} u d u\right)\\
&=\frac{1}{12} \cos ^{3} u \sin u+\frac{1}{4}\left(\frac{\sin u \cos u}{2}+\frac{1}{2} \int \cos ^{\theta} u d u\right) \\
&=\frac{1}{12} \cos ^{3} u \sin u+\frac{1}{4}\left(\frac{\sin 2 u}{4}+\frac{u}{2}\right)+c\\
&=\frac{1}{12} \cos ^{3}(3 x+2) \sin (3 x+2)+\frac{1}{8}(3 x+2)+\frac{1}{16} \sin (6 x+4)+c
\end{align*}
