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Math Help - Find the Indefinite Integral (cosx/1-cos^2x)dx

  1. #1
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    Find the Indefinite Integral (cosx/1-cos^2x)dx

    Find the indefinite Integral (cosx/1-cos^2x)dx.


    Thanks so much for your help in avance!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Safety33 View Post
    Find the indefinite Integral (cosx/1-cos^2x)dx.


    Thanks so much for your help in avance!
    1-\cos^2(x)=\sin^2(x)

    So we have

    \int\frac{\cos(x)}{\sin^2(x)}dx

    Now let u=\sin(x)\Rightarrow{du=\cos(x)}

    So now we have

    \int\frac{du}{u^2}

    You can go from there, yeah?
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  3. #3
    Super Member wingless's Avatar
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    \int \frac{\cos x}{1-\cos^2 x}~dx

    The most known trigonometric identity: \sin^2 x + \cos^2 x = 1.

    So 1-\cos^2 x = \sin^2 x.

    \int \frac{\cos x}{\sin^2 x}~dx

    Substitute u=\sin x...
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  4. #4
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    This is actually a very easy one.

    Rewrite as:

    \int cot(x)csc(x)dx

    Now, let u=csc(x), \;\ du=-cot(x)csc(x)dx

    Now you get:

    -\int du

    =-u

    =-csc(x)

    Add a constant C if you need to.
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  5. #5
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    Im coming out with 1/sinx. Is that incorrect?

    I have U^-1, so 1/U=1/sinx
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Safety33 View Post
    Im coming out with 1/sinx. Is that incorrect?

    I have U^-1, so 1/U=1/sinx
    \int\frac{du}{u^2}=\frac{{\color{red}{-1}}}{u}
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  7. #7
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    Where does the -1 come from? Im sorry Im struggling so much with this.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Safety33 View Post
    Where does the -1 come from? Im sorry Im struggling so much with this.
    \int{u^n}du=\frac{u^{n+1}}{n+1}\quad\text{provided that }n\ne{-1}

    So we have

    \int\frac{du}{u^2}=\int{u^{-2}du}=\frac{u^{-1}}{-1}=-u^{-1}=\frac{-1}{u}
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  9. #9
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    Im such and idiot! Thank you very much for walking me through that.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Safety33 View Post
    Im such and idiot! Thank you very much for walking me through that.
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  11. #11
    Eater of Worlds
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    \frac{-1}{sin(x)}=-csc(x).

    Same thing. You are good.
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  12. #12
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    Quote Originally Posted by Mathstud28 View Post

    \int\frac{du}{u^2}=\frac{{\color{red}{-1}}}{u}+{\color{blue}k}
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    \frac{-1}{u}+{\color{red}\bold{C}}

    ...who does k anyways..
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