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TheBabe

Little Review: what is the integral of sin^3(2x)

  • 2 years ago
  • 2 years ago

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  1. TheBabe
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    all right, \[\sin^3(2x)\]

    • 2 years ago
  2. TheBabe
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    seems like i need to do substitution rule, then use trig identities?

    • 2 years ago
  3. LeoMessi
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    right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

    • 2 years ago
  4. TheBabe
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    \[\int\limits_{}^{}\sin^3 x\]

    • 2 years ago
  5. LeoMessi
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    forgot the dx ;), so that is equal to \[(1/2)\int\limits_{}^{}\sin^3(u)du\]

    • 2 years ago
  6. LeoMessi
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    \[sin^2x=1-cos^2x\], so we'll have \[(1/2)\int\limits_{}^{}\sin^3(u)du\] = \[(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du\] = \[(1/2)\int\limits_{}^{}(1-sin^2(u))sin(u)du\]

    • 2 years ago
  7. LeoMessi
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    so we can apply substitution rule again for \[v=\sin(u)\]

    • 2 years ago
  8. LeoMessi
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    \[dv = \cos(u)du\], so \[(1/2)\int\limits_{}^{}(1-v^2)dv\]

    • 2 years ago
  9. TheBabe
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    wait, shouldn't it be 1-cos^2(u) above?

    • 2 years ago
  10. LeoMessi
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    ahhh, yesss... sorry: \[(1/2)\int\limits_{}^{}(1-cos^2(u))sin(u)du\]

    • 2 years ago
  11. TheBabe
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    so v=cos(u), which is dv=-sin(u)du

    • 2 years ago
  12. TheBabe
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    i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du = (-1)*(1/2) \int\limits_{}^{} (1-v^2)dv = (-1)(1/2) [ v-(1/3)v^3 + C]\]

    • 2 years ago
  13. TheBabe
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    i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du\] = \[(-1)*(1/2) \int\limits_{}^{} (1-v^2)dv\] = \[(-1)(1/2) [ v-(1/3)v^3 + C]\]

    • 2 years ago
  14. TheBabe
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    so plug back in and get v = cos u --> cos (2x), \[(-1)(1/2) [ v-(1/3)v^3 + C] = (-1)(1/2) [ \cos(2x) - (1/3)\cos^3(2x) + C]\] \[ = (-1)(1/2) [ cos(2x) - (1/3)cos^3(2x) + C]\]

    • 2 years ago
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