Answer
\begin{aligned}
\int \cos ^{3} 4 x \ d x =\frac{1}{4} \sin 4 x-\frac{1}{12} \sin ^{3} 4 x+C
\end{aligned}
Work Step by Step
Given $$ \int \cos ^{3} 4 x \ d x $$
So, we have
\begin{aligned}
I&=
\int \cos ^{3} 4 x \ d x\\
&=\int \cos ^{2} 4 x \cos 4 x\ d x\\
&=\frac{1}{4} \int\left(1-\sin ^{2} 4 x\right) \cos 4 x \cdot 4 d x\\
&=\frac{1}{4} \int \cos 4 x \cdot 4 d x-\frac{1}{4} \int \sin ^{2} 4 x \cos 4 x \cdot 4 d x\\
&=\frac{1}{4} \sin 4 x-\frac{1}{12} \sin ^{3} 4 x+C
\end{aligned}
