anonymous
  • anonymous
Integrate sqrt{5sin^{2}(2t)} from 0 to pi
Mathematics
jamiebookeater
  • jamiebookeater
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watchmath
  • watchmath
\(\sin^2(t)=\frac{1}{2}(1-cos(2t))\) So \(\int 5\sin^2(2t)\,dt=\int \frac{5}{2}(1-\cos(4t))\,dt=\frac{5}{2}t-\frac{5}{8}\sin(4t)+C\)
watchmath
  • watchmath
Oh sorry this is a definite integral... so just plugin the limits of integration :).
anonymous
  • anonymous
How did you eliminate the square root?

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watchmath
  • watchmath
I am sorry, I didn't see the square root ... :)
watchmath
  • watchmath
Ok, then it is simply the integral \[\sqrt{5}\int_0^\Pi \sin(2t)\,dt\] I believe you can do this integral now :).
watchmath
  • watchmath
Sorry ... I made a mistake
watchmath
  • 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 \[ \int_0^{\pi/2} \sqrt{5}\sin(2t)\,dt-\int_{\pi/2}^{\pi} \sqrt{5}\sin(2t)\,dt\]
anonymous
  • anonymous
Thank you so much, this is the part I was missing.

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