$\int (2 + 2 \cos x)^{1 / 2} dx$

By Anonymous on Thursday, March 8, 2001 - 06:33 pm :

How do you integrate $\int (2 + 2 \cos x)^{1 / 2} dx$ ? Thanks

By Dan Goodman (Dfmg2) on Thursday, March 8, 2001 - 06:47 pm :

You have to notice that

\cos (2 x) = \cos^{2} (x) - \sin^{2} (x) = 2 \cos^{2} (x) - 1

So $1 + \cos (2 x) = 2 \cos^{2} (x)$ .

So your integral is just $\int (2 (1 + \cos (x {))}^{1 / 2} dx = \int (4 \cos^{2} (x / 2 {))}^{1 / 2} dx = 2 \int \cos (x / 2) dx = 4 \sin (x / 2) + C$ where $C$ is a constant.

The things to remember are that $\cos (2 x) = \cos^{2} (x) - \sin^{2} (x)$ , $\sin (2 x) = 2 \cos (x) \sin (x)$

You should also try to remember that $1 + \cos (2 x)$ can be written as a multiple of $\cos^{2} (x)$ and $1 - \cos (2 x)$ is a multiple of $\sin^{2} (x)$ , and then just quickly work out the exact form when you need it.

By Anonymous on Thursday, March 8, 2001 - 07:19 pm :

Thanks for those tips Dan! ;)