Answer
\[ = \frac{1}{7}{\sec ^7}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\sin x{{\sec }^8}xdx} \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
\int_{}^{} {\sin x\left( {\frac{1}{{\cos x}}} \right)\sec x{{\sec }^6}xdx} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{\sin x}}{{\cos x}}} \right)\,\left( {\sec x} \right)\,\left( {{{\sec }^6}x} \right)} dx \hfill \\
\hfill \\
= \int_{}^{} {{{\sec }^6}x\,\left( {\sec x} \right)\,\left( {\tan x} \right)dx} \hfill \\
\hfill \\
set\,\,u = \sec x\,\,then\,\,du = \sec x\tan xdx \hfill \\
\hfill \\
= \int {{u^6}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
\frac{{{u^7}}}{7} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,u = \sec x \hfill \\
\hfill \\
= \frac{1}{7}{\sec ^7}x + C \hfill \\
\end{gathered} \]
