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