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anonymous

  • 5 years ago

How do I find the integral of sin(x)^2

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  1. anonymous
    • 5 years ago
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    try integration by parts

  2. anonymous
    • 5 years ago
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    u = sinx dv = sinx du = cosx v = cosx sinxcosx - \[\int\limits_{?}^{?} \cos^2x\]

  3. anonymous
    • 5 years ago
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    i don't see how thats going to get me anywhere

  4. anonymous
    • 5 years ago
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    sorry is it \[\sin ^{2}x\] or \[\sin x ^{2}\]???

  5. anonymous
    • 5 years ago
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    the first.

  6. anonymous
    • 5 years ago
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    \[\sin ^2x\]

  7. anonymous
    • 5 years ago
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    \[\cos ^{2}x=1-\sin ^{2}x\] \[\int\limits_{?}^{?}\sin ^{2}xdx=-\sin x \cos x +\int\limits_{?}^{?}(1-\sin ^{2}x)dx\]

  8. anonymous
    • 5 years ago
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    therefore, the last integral of sin^2(x) can be put to the left hand side.

  9. anonymous
    • 5 years ago
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    giving integral (sin^2(x))=(-sin(x)cos(x)+x)/2

  10. anonymous
    • 5 years ago
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    would the two integrals divided by each other reduce to -1 though? so it'd be =sinxcosx +1 in the end?

  11. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?}\sin^2x\] \[\div\] \[\int\limits_{}^{} 1-\sin^2x\] = -1?

  12. anonymous
    • 5 years ago
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    or is that just improper math? I'm not sure, I dont remember ever dividing integrals.

  13. anonymous
    • 5 years ago
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    oh wait, I wouldnt be dividng them though, I'd subtract, though I still think I'm missing something where does the 2 in -sinxcosx/2 come from?

  14. anonymous
    • 5 years ago
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    no, because you need to add integral (sin^2(x)) to both sides, not dividing them, giving 2 times integral (sin^2(x)) on the LHS and cancelling it on the RHS

  15. anonymous
    • 5 years ago
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    Oh, okay, I get it all now I forgot about separating integrals and what not. This all makes sense.

  16. anonymous
    • 5 years ago
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    My AP calc test is on wednesday, so, reviewing all this stuff is really helpful, but a pain in the retriceat the same time.

  17. anonymous
    • 5 years ago
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    good luck for wednesday.

  18. anonymous
    • 5 years ago
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    thanks.

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