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