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Math Help - Integrals by parts and substitution

  1. #1
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    Integrals by parts and substitution

    I'm not understanding what to do to the first problem after I cancel out the cscx in the top and bottom. The 2nd and 3rd problems are still kind of confusing.

    1) Int (cscx cotx)/((csc(x))^5) dx

    2) Int (dx)/(sqrt(1-16x^2)

    3) Int ((x^3)-(3x^2)+3x)/(x-1) dx
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  2. #2
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    Quote Originally Posted by rocsfinest762 View Post
    I'm not understanding what to do to the first problem after I cancel out the cscx in the top and bottom. The 2nd and 3rd problems are still kind of confusing.

    1) Int (cscx cotx)/((csc(x))^5) dx

    let u = \csc{x} ... du = -csc{x}\cot{x} \, dx

    2) Int (dx)/(sqrt(1-16x^2)

    integral is an arcsin ...

    \frac{1}{4} \int \frac{4}{\sqrt{1 - (4x)^2}} \, dx

    3) Int ((x^3)-(3x^2)+3x)/(x-1) dx

    do the long division
    .
    Last edited by skeeter; March 16th 2009 at 04:57 AM. Reason: edit ... fixed typo with du
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  3. #3
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    Integral

    Hello rocsfinest762
    Quote Originally Posted by rocsfinest762 View Post
    ...1) Int (cscx cotx)/((csc(x))^5) dx...
    I = \int\frac{\csc x \cot x}{\csc^5x}\,dx


    Now \csc x \cot x = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}=\frac{\cos x}{\sin^2x}

    and \csc^5x = \frac{1}{\sin^5x}

    \Rightarrow I = \int\frac{\cos x\sin^5x}{\sin^2x}\,dx

    = \int\cos x\sin^3x\,dx

    Let \sin x = u. Then \cos x \,dx =du

    I think you can complete it now.

    Grandad
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