Answer
\[ = \frac{1}{{20}}\sin 20x - \frac{1}{{20}}\,\left( {\frac{{\sin 20x}}{3}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\cos }^3}20xdx} \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
= \int_{}^{} {{{\cos }^2}20x \cdot \cos 20xdx} \hfill \\
\hfill \\
where \hfill \\
\hfill \\
{\cos ^2}20x = 1 - {\sin ^2}20x \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {1 - {{\sin }^2}20x} \right)\cos 20xdx} \hfill \\
\hfill \\
distribute \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\cos 20x} \right)dx} - \int_{}^{} {{{\sin }^2}20x\cos 20xdx} \hfill \\
\hfill \\
rewriting \hfill \\
\hfill \\
= \frac{1}{{20}}\int_{}^{} {\cos 20x\,\left( {20} \right)dx} - \frac{1}{{20}}\int_{}^{} {{{\sin }^2}20x\cos 20x\,\left( {20} \right)dx} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{{20}}\sin 20x - \frac{1}{{20}}\,\left( {\frac{{\sin 20x}}{3}} \right) + C \hfill \\
\end{gathered} \]