Answer
$$\frac{1}{36} \cos ^{3}(9 x-2) \sin (9 x-2)+\frac{1}{24} \cos (9 x-2) \sin (9 x-2)+\frac{(9 x-2)}{24}+C$$
Work Step by Step
Given$$\int \cos ^{4}(9 x-2) d x$$
Let $u= 9x-2\to du=9dx$ and use
$$\int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$$
Then
\begin{aligned} \int \cos ^{4}(9 x-2) d x &=\frac{1}{9} \int \cos ^{4} u d u \\ &=\frac{1}{9}\left[\frac{\cos ^{4-1} u \sin u}{4}+\frac{4-1}{4} \int \cos ^{4-2} u d u\right] \\ &=\frac{1}{9}\left[\frac{\cos ^{3} u \sin u}{4}+\frac{3}{4} \int \cos ^{2} u d u\right] \\ &=\frac{1}{9}\left[\frac{\cos ^{3} u \sin u}{4}+\frac{3}{4}\left[\frac{\cos u \sin u}{2}+\frac{2-1}{2} \int \cos ^{2-2} u d u\right]\right] \\ &=\frac{1}{9}\left[\frac{\cos ^{3} u \sin u}{4}+\frac{3}{4}\left[\frac{\cos u \sin u}{2}+\frac{1}{2} \int 1 d u\right]\right] \\ &\left.=\frac{\cos ^{3} u \sin u}{36}\left[\frac{\cos u \sin u}{2}+\frac{1}{2} \int 1 d u\right]\right] \\ &=\frac{1}{36} \cos ^{3} u \sin u+\frac{1}{12 \cdot 2}[\cos u \sin u+u]+C \\ &=\frac{1}{36} \cos ^{3}(9 x-2) \sin (9 x-2)+\frac{1}{24} \cos (9 x-2) \sin (9 x-2)+\frac{(9 x-2)}{24}+C \end{aligned}
