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

1. watchmath

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

2. watchmath

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

3. anonymous

How did you eliminate the square root?

4. watchmath

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

5. watchmath

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

6. watchmath

Sorry ... I made a mistake

7. watchmath

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 $\int_0^{\pi/2} \sqrt{5}\sin(2t)\,dt-\int_{\pi/2}^{\pi} \sqrt{5}\sin(2t)\,dt$

8. anonymous

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