How to deal with multiplication inside of integral?

Question

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.

Answer 1

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.

Answer 2

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