Evaluate:

$\displaystyle \displaystyle \int^{\frac{\pi}{2}}_0 [\frac{d}{dx} (\sin(\frac{x}{2}) \cos(\frac{x}{3})]dx$

Here is what I did:

$\displaystyle \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})dx + \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})(-\sin(\frac{x}{3}))dx$ (using the product rule to evaluate d/dx and separating the resultant integrals)

$\displaystyle \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})dx - \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})\sin(\frac{x}{3})dx$ (just rewriting to pull out the minus sign)

Assuming that this is right so far, I don't know how to evaluate these two integrals. I'm thinking that maybe I can use FTC to simplify this process somehow?

Thanks.