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

1. 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?

2. 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 $x\ge 0$, -x is x< 0.

So |sin(2x)|= sin(2x) if $0\le 2x\le \pi$ and -sin(2x) if $\pi\le 2x\le 2\pi$ which is the same as
|sin(2x)|= sin(2x) if [tex]0\le x\le \pi/2[/itex] and -sin(2x) if $\pi/2\le x\le \pi$

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

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integration of sin2t

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