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anonymous

  • one year ago

Use a substitution to find.....

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  1. anonymous
    • one year ago
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    |dw:1437223854442:dw|

  2. anonymous
    • one year ago
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    So i choose \[u=\cos 4t\] Derrivative gives \[\frac{ du }{ dt }=-4\sin 4t\] Then Rewrite \[\sin 4t dt = \frac{ du }{ 4 } \] and finally i got to this \[\frac{ 1 }{ 4 } \int\limits_{}^{} \sin ^{2} u dt\]

  3. anonymous
    • one year ago
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    Have I done correctly so far or no? What/where do I need to do changes.

  4. mathmate
    • one year ago
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    You can do the substitution in two steps, p=4t, and followed by q=sin(p), or as you suggested, you can do it in one step: u=sin(4t). I suggest u=sin(4t) because eventually the integration is simpler than u=cos(4t). I will first work it out using your suggested substitution. u=cos(4t); du=-4sin(4t)dt then I=\(\int sin^2(4t)cos(4t)dt\) =-\(\int (1/4)sin(4t)udu\) =-(1/4)\(\int sqrt(1-u^2)udu\) which makes the integral more awkward than necessary. If we proceed with u=sin(4t), life will be a little easier, as you can see below: u=sin(4t), du=4cos(4t)dt, or cos(4t)dt = (1/4)du I=\(\int sin^2(4t)cos(4t)dt\) =(1/4)\(\int u^2du\) =\(\Large \frac{u^3}{12}\) =\(\Large\frac{sin^3(4t)}{12}\) Rule of thumb: use substitution u=sin(x) when the integrand is \(sin^n(x)cos(x)dx\) or use substitution u=cos(x) when the integrand is \(cos^n(x)sin(x)dx\)

  5. mathmate
    • one year ago
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    @Xlegalize

  6. anonymous
    • one year ago
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    thanks, I solved it. But yeah, I see were I did my wrongs. THanks for ur time.

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