## Calculus: Early Transcendentals (2nd Edition)

$= \frac{{ - \cos \,\left( {{x^{10}}} \right)}}{{10}} + C$
$\begin{gathered} \int_{}^{} {{x^9}\sin {x^{10}}\,\,dx} \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = {x^{10}}\,\,\,\,then\,\,\,du = 10{x^9}dx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {{x^9}\sin {x^{10}}\,\,dx} = \,\frac{1}{{10}} \cdot \int_{}^{} {10{x^9}\sin {x^{10}}\,dx} \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \frac{1}{{10}} \cdot \int_{}^{} {\sin u\,du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ \frac{1}{{10}}\,\left( { - \cos \,u} \right) + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = {x^{10}} \hfill \\ \hfill \\ = \frac{{ - \cos \,\left( {{x^{10}}} \right)}}{{10}} + C \hfill \\ \end{gathered}$
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