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anonymous
 5 years ago
Little Review: what is the integral of sin^3(2x)
anonymous
 5 years ago
Little Review: what is the integral of sin^3(2x)

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

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

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

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

anonymous
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0so we can apply substitution rule again for \[v=\sin(u)\]

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

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

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

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

anonymous
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0so plug back in and get v = cos u > cos (2x), \[(1)(1/2) [ v(1/3)v^3 + C] = (1)(1/2) [ \cos(2x)  (1/3)\cos^3(2x) + C]\] \[ = (1)(1/2) [ cos(2x)  (1/3)cos^3(2x) + C]\]
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