# Integration problem

• Aug 31st 2010, 07:42 PM
RU2010
Integration problem
I'm having trouble integrating this function:

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

• Aug 31st 2010, 07:48 PM
mr fantastic
Quote:

Originally Posted by RU2010
I'm having trouble integrating this function:

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

Note that $\displaystyle \cos(4t) = 2 \cos^2 (2t) - 1$. See what you can do with that.
• Aug 31st 2010, 08:07 PM
RU2010
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 PM
Prove It
Better yet, note that $\displaystyle \cos{(4t)} = 1 - 2\sin^2{(2t)}$.
• Aug 31st 2010, 08:25 PM
RU2010
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 PM
RU2010
Scrap that substition, it will not work. Which method of integration do I need to look at?
• Aug 31st 2010, 08:46 PM
RU2010
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 PM
pickslides
$\displaystyle \int \sqrt{4 \sin^2{2t}}dt$

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

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

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

Its a gimmie from here!