Answer
$$\frac{1}{4} \cos ^{3} y \sin y+\frac{3}{8} \cos y \sin y+\frac{3}{8} y+C$$
Work Step by Step
Use
$$\int \cos ^{n} x d x=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x$$
We get
\begin{aligned} \int \cos ^{4} y d y &=\frac{1}{4} \cos ^{4-1} y \sin y+\frac{4-1}{4} \int \cos ^{4-2} y d y \\ &=\frac{1}{4} \cos ^{3} y \sin y+\frac{3}{4} \int \cos ^{2} y d y \\ &=\frac{1}{4} \cos ^{3} y \sin y+\frac{3}{4}\left[\frac{1}{2} \cos ^{2-1} y \sin y+\frac{2-1}{1} \int \cos ^{2-2} y d y\right] \\ &=\frac{1}{4} \cos ^{3} y \sin y+\frac{3}{4}\left[\frac{1}{2} \cos y \sin y+\frac{1}{2} y\right]+C \\ &=\frac{1}{4} \cos ^{3} y \sin y+\frac{3}{8} \cos y \sin y+\frac{3}{8} y+C \end{aligned}
