A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Little Review: what is the integral of sin^3(2x)
anonymous
 4 years ago
Little Review: what is the integral of sin^3(2x)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all right, \[\sin^3(2x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0seems like i need to do substitution rule, then use trig identities?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\sin^3 x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forgot the dx ;), so that is equal to \[(1/2)\int\limits_{}^{}\sin^3(u)du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[sin^2x=1cos^2x\], so we'll have \[(1/2)\int\limits_{}^{}\sin^3(u)du\] = \[(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du\] = \[(1/2)\int\limits_{}^{}(1sin^2(u))sin(u)du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we can apply substitution rule again for \[v=\sin(u)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[dv = \cos(u)du\], so \[(1/2)\int\limits_{}^{}(1v^2)dv\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait, shouldn't it be 1cos^2(u) above?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahhh, yesss... sorry: \[(1/2)\int\limits_{}^{}(1cos^2(u))sin(u)du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so v=cos(u), which is dv=sin(u)du

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1\cos^2(u))\sin(u)du = (1)*(1/2) \int\limits_{}^{} (1v^2)dv = (1)(1/2) [ v(1/3)v^3 + C]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1\cos^2(u))\sin(u)du\] = \[(1)*(1/2) \int\limits_{}^{} (1v^2)dv\] = \[(1)(1/2) [ v(1/3)v^3 + C]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so plug back in and get v = cos u > cos (2x), \[(1)(1/2) [ v(1/3)v^3 + C] = (1)(1/2) [ \cos(2x)  (1/3)\cos^3(2x) + C]\] \[ = (1)(1/2) [ cos(2x)  (1/3)cos^3(2x) + C]\]
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.