# antiderivative of $\frac{1}{4}\sin(2x)$

I am trying to find the antiderivative of $\frac{1}{4}\sin(2x)$ 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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$\frac{1}{2}cos(2x)$ but not sure how to get it – soniccool Dec 13 '12 at 2:10
Make a guess: $-\cos(2x)$ seems reasonable. Check by differentiating: ${d\over dx}(-\cos (2x))=2\sin(2x)$. Doesn't quite work: so modify the original guess: ${1\over8}(-\cos 2x)+C$ is the answer. – David Mitra Dec 13 '12 at 2:25

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

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

Let $u = 2x, du = 2 \ dx \implies \frac{du}{2} = dx$.

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

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

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.

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

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: $$\int \sin x \ dx = -\cos x + C$$ $$\int \cos x \ dx = \sin x + C$$

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.

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The answer is $\frac{1}{2}cos(2x)$ but im not sure how to get it. – soniccool Dec 13 '12 at 2:10
That answer is correct only if your original was $-\sin(2x)$. – Joe Dec 13 '12 at 2:12

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

Then you can write

$$\int \frac{1}{4}\sin(2x)\mathrm{d}x = \frac{1}{2}\int \sin(x)\cos(x)\mathrm{d}x \\$$ Set $u=\cos(x)$ then $du=-\sin(x)dx$. 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 $u=\cos(x)$ we get $$\int \frac{1}{4}\sin(2x)\mathrm{d}x = -\frac{1}{4} \cos^2(x) + C$$ This looks very different but you can see that it's equivalent to the first answer given to your question by remembering the following: $$\cos^2(x) = 1 - \sin^2(x)$$ and $$\sin^2(x) = \frac{1-\cos(2x)}{2}$$ and finally keeping in mind that $C$ of the two answers is not the same.

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