Integrating $3\sin(x/2)$.

$$\int 3\sin\left(\frac{x}{2}\right)dx$$

can't figure this one out! I'm not sure if I'm supposed to substitute or not?

Here's where I'm at...

$$3\int\sin\left(\frac{x}{2}\right)dx$$ $$u = \frac{x}{2}$$ $$du = \frac{1}{2}dx$$ $$dx = 2du$$ $$3\int\sin\left(u\right)2du$$

and then...

$$-6cos\left(\frac{u^2}{2}\right)$$ thats not right is it..?

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Hint. Set $u=\frac{x}{2}$ and do a substitution. –  Arturo Magidin Jul 24 '12 at 3:24
I actually just tried that before I asked this and it turned out ugly! :[ –  Kudla69 Jul 24 '12 at 3:24
Then you are doing things wrong. It doesn't "turn ugly". Post your work and we'll tell you where you are messing up. –  Arturo Magidin Jul 24 '12 at 3:26
Is this homework? –  Ian Mateus Jul 24 '12 at 3:27
And how is that "ugly"? The factor of $2$ is multiplying, so you can pull it out exactly the same way you pulled out the $3$. And surely you know what $\int\sin(u)\,du$ is! –  Arturo Magidin Jul 24 '12 at 3:30
Let $x=2t\implies dx=2dt$. Hence your problem becomes $2\int 3\sin(t)dt=-6\cos(t)+c=-6\cos(x/2)+c$ where $c$ is a constant.