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I have an undefined integral like this: \begin{aligned} \ \int x^3 \cdot \sin(4+9x^4)dx \end{aligned}

I have to integrate it and I have no idea where to start. I have basic formulas for integrating but I need to split this equation into two or to do something else.

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    thanks!!! How could I have been so blind?2012-02-19

2 Answers 2

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Note that (4+9x^4)' = 36x^3

So that your integral becomes

$\int x^3 \sin(4+9x^4)dx$

$\dfrac{1}{36}\int 36x^3 \sin(4+9x^4)dx$

$\dfrac{1}{36}\int \sin u du$

Which you can easily solve.

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Hint: $\begin{eqnarray*} (\cos (u(x)))^{\prime } &=&-\sin (u(x))u^{\prime }(x)\qquad\text{(by the chain rule)} \\ &\Leftrightarrow &\int \sin (u(x))u^{\prime }(x)dx=-\cos (u(x))+\text{Const.} \end{eqnarray*}$