# Antiderivative of 1 4 </mfrac> sin &#x2061;<!-- ⁡ --> ( 2 x )

Antiderivative of $\frac{1}{4}\mathrm{sin}\left(2x\right)$
I am trying to find the antiderivative of $\frac{1}{4}\mathrm{sin}\left(2x\right)$ but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?
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Blaze Frank
Step 1
The beauty of trig is that you can get wildly different looking results depending on how you start. For example take identify: $\mathrm{sin}\left(2x\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$
Then you can write $\int \frac{1}{4}\mathrm{sin}\left(2x\right)\mathrm{d}x=\frac{1}{2}\int \mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)\mathrm{d}x\phantom{\rule{0ex}{0ex}}$.
Set $u=\mathrm{cos}\left(x\right)$ then $du=-\mathrm{sin}\left(x\right)dx$. So now we solve the indefinite integral: $\frac{1}{2}\int -u\mathrm{d}u=-\frac{1}{4}{u}^{2}+C$
Substituting back in for $u=\mathrm{cos}\left(x\right)$ we get
$\int \frac{1}{4}\mathrm{sin}\left(2x\right)\mathrm{d}x=-\frac{1}{4}{\mathrm{cos}}^{2}\left(x\right)+C$
This looks very different but you can see that it's equivalent to the first answer given to your question by remembering the following:
${\mathrm{cos}}^{2}\left(x\right)=1-{\mathrm{sin}}^{2}\left(x\right)$
and ${\mathrm{sin}}^{2}\left(x\right)=\frac{1-\mathrm{cos}\left(2x\right)}{2}$ and finally keeping in mind that C of the two answers is not the same.

Arraryeldergox2
Step 1

Let .

$I=\frac{1}{8}\cdot -\mathrm{cos}u+C$
$I=-\frac{1}{8}\mathrm{cos}\left(2x\right)+C$
Why is this correct? We can take the derivative of our answer to see we get our original function back and that this and our primitive are equal up to a constant.
$\frac{d}{dx}\left(-\frac{1}{8}\mathrm{cos}\left(2x\right)\right)=-\frac{1}{8}\cdot -\mathrm{sin}\left(2x\right)\cdot 2=\frac{1}{4}\mathrm{sin}\left(2x\right)$
It is helpful to note these few facts to answer your question about what to do when you have sin or cos in an integrand:

Note that this just involves x, but you can still do your normal u-substitutions as necessary to get it in the form above, as I did.