How to deal with multiplication inside of integral?
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.
$\endgroup$ 22 Answers
$\begingroup$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.
$\endgroup$ 0 $\begingroup$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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