I'm having trouble integrating this function:

$\displaystyle \int \sqrt{2-2\cos{4t}} dt$

So far I've tried substitution and that didn't work out so well. Please help!

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- Aug 31st 2010, 07:42 PMRU2010Integration problem
I'm having trouble integrating this function:

$\displaystyle \int \sqrt{2-2\cos{4t}} dt$

So far I've tried substitution and that didn't work out so well. Please help! - Aug 31st 2010, 07:48 PMmr fantastic
- Aug 31st 2010, 08:07 PMRU2010
Ok, I made the double angle substitution, but I'm still drawing a blank on how to get this one started. Which method of integration should I use?

- Aug 31st 2010, 08:07 PMProve It
Better yet, note that $\displaystyle \cos{(4t)} = 1 - 2\sin^2{(2t)}$.

- Aug 31st 2010, 08:25 PMRU2010
Ok, I've used the trig identities to change the integral to this:

$\displaystyle \int \sqrt{4 \sin^2{2t}}dt$

I think what is throwing me off is the fact that this function is under the radial. At this point, would substitution work if I use $\displaystyle u= \sin^2{2t}$? Is this the correct integration method to use? - Aug 31st 2010, 08:32 PMRU2010
Scrap that substition, it will not work. Which method of integration do I need to look at?

- Aug 31st 2010, 08:46 PMRU2010
I think I might have found the solution. The problem is actually a definite integral problem with the lower limit of integration as 0 and the upper limit of integration as pi/2.

I've worked the problem out and came to a solution of -2. Is this correct? - Aug 31st 2010, 08:52 PMpickslides
$\displaystyle \int \sqrt{4 \sin^2{2t}}dt$

$\displaystyle \int \sqrt{2^2 \sin^2{2t}}dt$

$\displaystyle \int \sqrt{(2 \sin{2t})^2}dt$

$\displaystyle \int 2 \sin{2t}dt$

Its a gimmie from here!