# 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! :[ – user69 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
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
this particular problem is what I am struggling with. I missed a lesson so I'm pretty much learning substitution and integration at the same time. The integration isn't hard, It's just that this problem confused me because of the trig function with the fraction inside – user69 Jul 24 '12 at 3:39

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.