Answer
\[ = - \frac{{{{\cos }^9}{x^2}}}{{18}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {x\sin {x^2}{{\cos }^8}{x^2}dx} \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
= - \frac{1}{2}\int_{}^{} {\,\left( {{{\cos }^8}{x^2}} \right)\,\left( { - \sin {x^2}} \right)\,\left( {2x} \right)dx} \hfill \\
\hfill \\
integrate\, \hfill \\
\hfill \\
use\,\,\int_{}^{} {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C \hfill \\
\hfill \\
= - \frac{1}{2}\,\left( {\frac{{{{\cos }^9}{x^2}}}{9}} \right) + C \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= - \frac{{{{\cos }^9}{x^2}}}{{18}} + C \hfill \\
\end{gathered} \]