A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Integrate sqrt{5sin^{2}(2t)} from 0 to pi
anonymous
 5 years ago
Integrate sqrt{5sin^{2}(2t)} from 0 to pi

This Question is Closed

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(\sin^2(t)=\frac{1}{2}(1cos(2t))\) 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
 5 years ago
Best ResponseYou've already chosen the best response.1Oh sorry this is a definite integral... so just plugin the limits of integration :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you eliminate the square root?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I am sorry, I didn't see the square root ... :)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Ok, then it is simply the integral \[\sqrt{5}\int_0^\Pi \sin(2t)\,dt\] I believe you can do this integral now :).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Sorry ... I made a mistake

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Over the interval \([0,\pi]\) the function \(\sin(2t\) is above the \(taxis\) 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you so much, this is the part I was missing.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.