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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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9  
Remember that integration is basically undoing differentiation. There are two rules from differentiation that result in products of things: the chain rule and the product rule. These two rules give rise to u-substitution and integration by parts. Generally you want to see if you can find a solution by u-substitution before trying integration by parts, since it is a bit easier. So here's a hint: do you see a function and its derivative somewhere? The hard part of this problem isn't the product by the way, it's the stuff inside the sin. That's going to be what you want to check. –  Chris Janjigian Feb 19 '12 at 21:41
    
thanks!!! How could I have been so blind? –  Povylas Feb 19 '12 at 21:45

2 Answers 2

up vote 3 down vote accepted

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

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