anonymous
  • anonymous
Little Review: what is the integral of sin^3(2x)
OCW Scholar - Single Variable Calculus
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
all right, \[\sin^3(2x)\]
anonymous
  • anonymous
seems like i need to do substitution rule, then use trig identities?
anonymous
  • anonymous
right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

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anonymous
  • anonymous
\[\int\limits_{}^{}\sin^3 x\]
anonymous
  • anonymous
forgot the dx ;), so that is equal to \[(1/2)\int\limits_{}^{}\sin^3(u)du\]
anonymous
  • anonymous
\[sin^2x=1-cos^2x\], so we'll have \[(1/2)\int\limits_{}^{}\sin^3(u)du\] = \[(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du\] = \[(1/2)\int\limits_{}^{}(1-sin^2(u))sin(u)du\]
anonymous
  • anonymous
so we can apply substitution rule again for \[v=\sin(u)\]
anonymous
  • anonymous
\[dv = \cos(u)du\], so \[(1/2)\int\limits_{}^{}(1-v^2)dv\]
anonymous
  • anonymous
wait, shouldn't it be 1-cos^2(u) above?
anonymous
  • anonymous
ahhh, yesss... sorry: \[(1/2)\int\limits_{}^{}(1-cos^2(u))sin(u)du\]
anonymous
  • anonymous
so v=cos(u), which is dv=-sin(u)du
anonymous
  • anonymous
i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du = (-1)*(1/2) \int\limits_{}^{} (1-v^2)dv = (-1)(1/2) [ v-(1/3)v^3 + C]\]
anonymous
  • anonymous
i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du\] = \[(-1)*(1/2) \int\limits_{}^{} (1-v^2)dv\] = \[(-1)(1/2) [ v-(1/3)v^3 + C]\]
anonymous
  • anonymous
so 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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