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Math Help - U-Substitution

  1. #1
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    U-Substitution

    I need help with the u-substituion. The first one is....

    (2cosx)/(1-cos^2x)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tbenne3 View Post
    I need help with the u-substituion. The first one is....

    (2cosx)/(1-cos^2x)
    Hint: 1-\cos^2x=\sin^2x. Now what is the substitution?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Hint: 1-\cos^2x=\sin^2x. Now what is the substitution?
    I really don't know..
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  4. #4
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    Quote Originally Posted by tbenne3 View Post
    I really don't know..
    better brush up on your basic identities ...

    \frac{2\cos{x}}{1 - \cos^2{x}} = \frac{2\cos{x}}{\sin^2{x}} = 2\csc{x}\cot{x}<br />
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    Quote Originally Posted by skeeter View Post
    better brush up on your basic identities ...

    \frac{2\cos{x}}{1 - \cos^2{x}} = \frac{2\cos{x}}{\sin^2{x}} = 2\csc{x}\cot{x}<br />
    k thanks
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  6. #6
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    Notice that if u = \sin{x} then du = \cos{x}\,dx.


    So \int{\frac{2\cos{x}}{\sin^2{x}}\,dx} = 2\int{u^{-2}\,du}

     = -2u^{-1} + C

     = -\frac{2}{\sin{x}}+ C

     = -2\csc{x} + C.
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  7. #7
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    You can also let u = 1-\cos^2{x}, then \dfrac{du}{dx} = -2\cos{x}\sin{x} \Rightarrow {dx} = \dfrac{du}{-2\cos{x}\sin{x}}<br />
so then \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \int\dfrac{2\cos{x}}{u\left(-2\cos{x}\sin{x}\right)}\;{du} = -\int\dfrac{1}{u\sin{x}}\;{du} . Now since u = 1-\cos^2{x} = \sin^2{x}, then \sin{x} = \sqrt{u}. So we have: -\int\dfrac{1}{u\sin{x}}\;{du}= -\int\dfrac{1}{u\sqrt{u}}\;{du} = -\int\dfrac{1}{\sqrt{u^3}}\;{du} = \dfrac{-2}{\sqrt{u}}+C \Rightarrow \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \dfrac{-2}{\sqrt{1-\cos^2{x}}}+C.<br />
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