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Math Help - Quick ? Integrals

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    Quick ? Integrals

    Please see Red


    Thanks!

    -qbkr21
    Attached Thumbnails Attached Thumbnails Quick ? Integrals-simplify.gif  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Please see Red


    Thanks!

    -qbkr21
    it's a trig identity:

    sin(2x) = 2sinxcosx

    Proof:
    we apply the double angle formula for sine

    sin2x = sin(x + x) = sinxcosx + sinxcosx = 2sinxcosx
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails Quick ? Integrals-integrate-3qq.gif  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Re:
    Congrats on your 5th post!

    the trick is to substitute 1 + cos^2x, so let u = 1 + cos^2x and things should work out nicely
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    Re:

    Hmmm...for some odd reason I though you integrated first with "u" before you substitute "u" back into the problem, is this not always the case?

    Ohh...I see I just substituted for cos(x) not 1+cos^2(x). I think that I made this a bit more difficult than it had to be.

    -qbkr21
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Hmmm...for some odd reason I though you integrated first with "u" before you substitute "u" back into the problem, is this not always the case?

    -qbkr21
    no, i meant at the begining. to get to the point you are now, it appears you let u = cos^2x. i'm saying go back and redo the problem from scratch using u = 1 + cos^2x

    this is my 17th post!!
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    Re:

    How did you tell it was 1+cos^2(x) from the beginning?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    How did you tell it was 1+cos^2(x) from the beginning?
    the whole point of substitution is to turn a complicated integral into a simple one. the simpler we can make things the better. i realized that substituting u = cosx or u = cos^2x only would not really get rid of the problem, since i'd have to make a second substitution, say v = u^2 to get rid of the resulting u in the numerator, or would have to deal with a 1 + u in the denominator (not really a big deal, but still). now if i substituted 1 + cos^2x, i would get rid of the constant and the cos^2x, the derivative of a constant is zero, so it doesn't matter in the long run. and the integral will simply be -1/u, and you can't get any better than that

    looking at the problem again, i saw that you just substituted cosx, but as you see, you can make life a lot simpler than that
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    Quote Originally Posted by qbkr21 View Post
    Re:
    Unless you are required to use a particular method when confronted with
    the integral:

    I = 2* integral u/(1+u^2) du

    the observation that 2u = d/du[1+u^2] should imeadiatly tell you that:

    I = ln(1+u^2) + c.

    The general idea is that:

    integral [df/dx]/f(x) dx = ln(f(x) + c

    RonL
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