Question

Integrate sqrt{5sin^{2}(2t)} from 0 to pi

Answer 1

\(\sin^2(t)=\frac{1}{2}(1-cos(2t))\)
So
\(\int 5\sin^2(2t)\,dt=\int \frac{5}{2}(1-\cos(4t))\,dt=\frac{5}{2}t-\frac{5}{8}\sin(4t)+C\)

Answer 2

Oh sorry this is a definite integral... so just plugin the limits of integration :).

Answer 3

How did you eliminate the square root?

Answer 4

I am sorry, I didn't see the square root ... :)

Answer 5

Ok, then it is simply the integral
\[\sqrt{5}\int_0^\Pi \sin(2t)\,dt\]
I believe you can do this integral now :).

Answer 6

Sorry ... I made a mistake

Answer 7

Over the interval \([0,\pi]\) the function \(\sin(2t\) is above the \(t-axis\) on \([0,\pi/2]\) and below the \(t\)-axis on \([\pi/2,\pi]\).
So \(\sqrt{5\sin^2{2t}}=\sqrt{5}\sin{2t}\) on \([0,\pi/2]\) and equal to \(-\sqrt{5}\sin(2t)\) on the interval \([\pi/2,\pi]\). So the integral si equal to
\[
\int_0^{\pi/2}
\sqrt{5}\sin(2t)\,dt-\int_{\pi/2}^{\pi} \sqrt{5}\sin(2t)\,dt\]

Answer 8

Thank you so much, this is the part I was missing.