Answer
\[ = 4{\tan ^5}x - \frac{{20{{\tan }^3}x}}{3} + 20\tan x - 20x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {20\,{{\tan }^6}xdx} \hfill \\
\hfill \\
write\,\,\,{\tan ^6}x\,\,as\,{\tan ^4}x{\tan ^2}x \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x{{\tan }^2}xdx} \hfill \\
\hfill \\
Use\,\,the\,identity\,\,{\tan ^2}x = {\sec ^2}x - 1 \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x\,\left( {{{\sec }^2}x - 1} \right)dx} \hfill \\
\hfill \\
multiply\, \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x{{\sec }^2}xdx - 20\int_{}^{} {{{\tan }^4}xdx} } \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}{{\sec }^2}xdx} - 20\int_{}^{} {{{\tan }^2}x{{\tan }^2}xdx} \hfill \\
\hfill \\
Use\,\,\,the\,\,\,identity\,\,\,\,{\tan ^2}x = {\sec ^2}x - 1 \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x{{\sec }^2}xdx} - 20\int_{}^{} {{{\tan }^2}x\,\left( {{{\sec }^2}x - 1} \right)dx} \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x{{\sec }^2}xdx - 20\int_{}^{} {{{\tan }^2}x{{\sec }^2}xdx + 20\int_{}^{} {{{\tan }^2}xdx} } } \hfill \\
\hfill \\
= 20\int_{}^{} {{{\tan }^4}x{{\sec }^2}xdx} - 20\int_{}^{} {{{\tan }^2}x{{\sec }^2}xdx} + 20\int_{}^{} {\,\left( {{{\sec }^2}x - 1} \right)dx} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= 20\,\left( {\frac{{{{\tan }^5}x}}{5}} \right) - 20\,\left( {\frac{{{{\tan }^3}x}}{3}} \right) + 20\tan x - 20x + C \hfill \\
\hfill \\
= 4{\tan ^5}x - \frac{{20{{\tan }^3}x}}{3} + 20\tan x - 20x + C \hfill \\
\end{gathered} \]
