How do I evaluate the integral of (sin(2t)^2)^(1/2) ?

I had to evaluate an integral from 0 to 2pi in which the integrand was sqrt(sin(2x)^2). I figured the square root of sin(2x) squared was just sin(2x). Evaluating the integral gave me 0, which makes sense since the function is symmetric. But this was "incorrect".

Apparently sqrt(sin(2x)^2) = |sin(2x)|. How would I express the integral of the absolute value of a function? (Please generalize your answer so that it explains cases for which the function is not symmetric.)

|sin(2x)| = sin(2x) and -sin(2x) so would the integral of |sin(2x)| from 0 to 2pi be the sum of the integral of sin(2x) and the integral of -sin(2x) from 0 to 2pi?

Re: How do I evaluate the integral of (sin(2t)^2)^(1/2) ?

It should be obvious that the sum of sin(2x) and -sin(2x) is 0 and so the sum of the integrals will be 0!

The basic definition of "absolute value" is that |x|= x if $\displaystyle x\ge 0$, -x is x< 0.

So |sin(2x)|= sin(2x) if $\displaystyle 0\le 2x\le \pi$ and -sin(2x) if $\displaystyle \pi\le 2x\le 2\pi$ which is the same as

|sin(2x)|= sin(2x) if [tex]0\le x\le \pi/2[/itex] and -sin(2x) if $\displaystyle \pi/2\le x\le \pi$

If you want to integrate from 0 to $\displaystyle 2\pi$, you will also need to use the fact that, sine is periodic with period $\displaystyle 2\pi$, sin(x) is positive for $\displaystyle 2\pi\< x< 3\pi$ and negative for $\displaystyle 3\pi< x< 4\pi$ so |sin(2x)|= sin(2x) for $\displaystyle \pi< x< 3\pi/2$ and |sin(2x)|= -sin(2x) for $\displaystyle 3\pi/2< x< 2\pi$