# Calc1.51

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$\int x^{2}\sin x^{3}\,dx\,$

We know how to do the integral of $\sin x\,$ so a good substitution to try would be $u=x^{3}\,$ because if it works out we will be left with $\sin u\,$ which we can integrate.

$u=x^{3}\,$

$du=3x^{2}dx\,$

$dx={\frac {du}{3x^{2}}}\,$

Substitute these into the integral to see if it changes the integral into a form we can integrate.

$\int x^{2}\sin x^{3}\,dx=\int x^{2}\sin u{\frac {du}{3x^{2}}}={\frac {1}{3}}\int \sin u\,du\,$

This is an integral we can compute.

${\frac {1}{3}}\int \sin u\,du=-{\frac {1}{3}}\cos(u)+C\,$

Finally, undo the substitution

$-{\frac {1}{3}}\cos(u)+C=-{\frac {1}{3}}\cos(x^{3})+C\,$