Summer Bradford

2022-06-25

Antiderivative of $\frac{1}{4}\mathrm{sin}(2x)$

I am trying to find the antiderivative of $\frac{1}{4}\mathrm{sin}(2x)$ but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?

I am trying to find the antiderivative of $\frac{1}{4}\mathrm{sin}(2x)$ but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?

Blaze Frank

Beginner2022-06-26Added 18 answers

Step 1

The beauty of trig is that you can get wildly different looking results depending on how you start. For example take identify: $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

Then you can write $\int \frac{1}{4}\mathrm{sin}(2x)\mathrm{d}x=\frac{1}{2}\int \mathrm{sin}(x)\mathrm{cos}(x)\mathrm{d}x\phantom{\rule{0ex}{0ex}}$.

Set $u=\mathrm{cos}(x)$ then $du=-\mathrm{sin}(x)dx$. 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 $u=\mathrm{cos}(x)$ we get

$\int \frac{1}{4}\mathrm{sin}(2x)\mathrm{d}x=-\frac{1}{4}{\mathrm{cos}}^{2}(x)+C$

This looks very different but you can see that it's equivalent to the first answer given to your question by remembering the following:

${\mathrm{cos}}^{2}(x)=1-{\mathrm{sin}}^{2}(x)$

and ${\mathrm{sin}}^{2}(x)=\frac{1-\mathrm{cos}(2x)}{2}$ and finally keeping in mind that C of the two answers is not the same.

The beauty of trig is that you can get wildly different looking results depending on how you start. For example take identify: $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

Then you can write $\int \frac{1}{4}\mathrm{sin}(2x)\mathrm{d}x=\frac{1}{2}\int \mathrm{sin}(x)\mathrm{cos}(x)\mathrm{d}x\phantom{\rule{0ex}{0ex}}$.

Set $u=\mathrm{cos}(x)$ then $du=-\mathrm{sin}(x)dx$. 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 $u=\mathrm{cos}(x)$ we get

$\int \frac{1}{4}\mathrm{sin}(2x)\mathrm{d}x=-\frac{1}{4}{\mathrm{cos}}^{2}(x)+C$

This looks very different but you can see that it's equivalent to the first answer given to your question by remembering the following:

${\mathrm{cos}}^{2}(x)=1-{\mathrm{sin}}^{2}(x)$

and ${\mathrm{sin}}^{2}(x)=\frac{1-\mathrm{cos}(2x)}{2}$ and finally keeping in mind that C of the two answers is not the same.

Arraryeldergox2

Beginner2022-06-27Added 10 answers

Step 1

$I:=\int \frac{1}{4}\mathrm{sin}(2x)\text{}dx$

$I=\frac{1}{4}\int \mathrm{sin}(2x)\text{}dx$

Let $u=2x,du=2\text{}dx\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{du}{2}=dx$.

$I=\frac{1}{4}\cdot \frac{1}{2}\int \mathrm{sin}(u)\text{}du$

$I=\frac{1}{8}\cdot -\mathrm{cos}u+C$

$I=-\frac{1}{8}\mathrm{cos}(2x)+C$

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.

$\frac{d}{dx}(-\frac{1}{8}\mathrm{cos}(2x))=-\frac{1}{8}\cdot -\mathrm{sin}(2x)\cdot 2=\frac{1}{4}\mathrm{sin}(2x)$

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:

$\int \mathrm{sin}x\text{}dx=-\mathrm{cos}x+C$

$\int \mathrm{cos}x\text{}dx=\mathrm{sin}x+C$

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.

$I:=\int \frac{1}{4}\mathrm{sin}(2x)\text{}dx$

$I=\frac{1}{4}\int \mathrm{sin}(2x)\text{}dx$

Let $u=2x,du=2\text{}dx\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{du}{2}=dx$.

$I=\frac{1}{4}\cdot \frac{1}{2}\int \mathrm{sin}(u)\text{}du$

$I=\frac{1}{8}\cdot -\mathrm{cos}u+C$

$I=-\frac{1}{8}\mathrm{cos}(2x)+C$

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.

$\frac{d}{dx}(-\frac{1}{8}\mathrm{cos}(2x))=-\frac{1}{8}\cdot -\mathrm{sin}(2x)\cdot 2=\frac{1}{4}\mathrm{sin}(2x)$

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:

$\int \mathrm{sin}x\text{}dx=-\mathrm{cos}x+C$

$\int \mathrm{cos}x\text{}dx=\mathrm{sin}x+C$

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.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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