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Math Help - Integration By Parts

  1. #1
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    Integration By Parts

    I'm having a lot of difficulty figuring out:

    csc5(x) dx

    My instructor said something about "Looping the Integral" if that helps at all.

    The instructor also gave us an example of:

    csc3(x) dx

    But i'm not sure how to go about the first one I gave.
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  2. #2
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    Re: Integration By Parts

    Hello, Preston019!

    I'm having a lot of difficulty figuring out: . \int\csc^5\!x\,dx

    My instructor said something about "Looping the Integral" if that helps at all.

    The instructor also gave us an example of: . \int \csc^3\!x\,dx

    But I'm not sure how to go about the first one I gave.

    Did you understand your instructor's example?
    If not, no wonder you can't solve this one.
    Maybe I'd better baby-step through it . . .

    We have: . I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)

    Integrate by parts: . \begin{Bmatrix}u &=& \csc x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}\csc x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}

    We have: . I \;=\;\text{-}\csc x\cot x - \int\csc x\cot^2\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \int\csc x(\csc^2\!x-1)\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \int(\csc^3\!x - \csc x)\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \underbrace{\int\csc^3\!x\,dx}_{\text{This is }I} + \int\csc x\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - I + \int\csc x\,dx

    . . . . . . . 2I \;=\;\text{-}\csc x\cot x + \int\csc x\,dx

    . . . . . . . 2I \;=\;\text{-}\csc x\cot x + \ln|\csc x - \cot x| + C

    . . . . . . . . I \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C

    Therefore: . \int\csc^3x\,dx \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We have: . I \;=\; \int\!\csc^5\!x\,dx \;=\;\int\!\csc^3\!x(\csc^2\!x\,dx)

    Integrate by parts: . \begin{Bmatrix}u &=& \csc^3\!x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}3\csc^3\!x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}

    We have: . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x(\csc^2\!x-1)\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int(\csc^5\!x - \csc^3\!x)\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\underbrace{\int\!\csc^5\!x\,dx}_{\text{This is }I} + 3\!\int\!\csc^3\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3I + 3\!\int\!\csc^3\!x\,dx

    . . . . . . . 4I \;=\;\text{-}\csc^3\!x\cot x + 3\!\int\!\csc^3\!x\,dx


    Since we know the value of \int\!\csc^3\!x\,dx, we can finish the problem . . .
    Thanks from Preston019
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  3. #3
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    Re: Integration By Parts

    Quote Originally Posted by Preston019 View Post
    I'm having a lot of difficulty figuring out:

    csc5(x) dx

    My instructor said something about "Looping the Integral" if that helps at all.

    The instructor also gave us an example of:

    csc3(x) dx

    But i'm not sure how to go about the first one I gave.
    \displaystyle \begin{align*} \int{\csc^5{x}\,dx} &= \int{\frac{1}{\sin^5{x}}\,dx} \\ &= \int{\frac{\sin{x}}{\sin^6{x}}\,dx} \\ &= -\int{\frac{-\sin{x}}{\left(1 - \cos^2{x} \right)^3}\,dx} \\ &= -\int{\frac{1}{\left(1 - u^2\right)^3}\,du} \textrm{ after making the substitution }u = \cos{x} \\ &= -\int{\frac{1}{(1 - u)^3(1 + u)^3}\,du} \end{align*}

    You should be able to solve this using Partial Fractions.
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  4. #4
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    Re: Integration By Parts

    Quote Originally Posted by Soroban View Post
    Hello, Preston019!


    Did you understand your instructor's example?
    If not, no wonder you can't solve this one.
    Maybe I'd better baby-step through it . . .

    We have: . I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)

    Integrate by parts: . \begin{Bmatrix}u &=& \csc x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}\csc x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}

    We have: . I \;=\;\text{-}\csc x\cot x - \int\csc x\cot^2\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \int\csc x(\csc^2\!x-1)\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \int(\csc^3\!x - \csc x)\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - \underbrace{\int\csc^3\!x\,dx}_{\text{This is }I} + \int\csc x\,dx

    . . . . . . . . I \;=\;\text{-}\csc x\cot x - I + \int\csc x\,dx

    . . . . . . . 2I \;=\;\text{-}\csc x\cot x + \int\csc x\,dx

    . . . . . . . 2I \;=\;\text{-}\csc x\cot x + \ln|\csc x - \cot x| + C

    . . . . . . . . I \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C

    Therefore: . \int\csc^3x\,dx \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We have: . I \;=\; \int\!\csc^5\!x\,dx \;=\;\int\!\csc^3\!x(\csc^2\!x\,dx)

    Integrate by parts: . \begin{Bmatrix}u &=& \csc^3\!x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}3\csc^3\!x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}

    We have: . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x(\csc^2\!x-1)\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int(\csc^5\!x - \csc^3\!x)\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3\underbrace{\int\!\csc^5\!x\,dx}_{\text{This is }I} + 3\!\int\!\csc^3\!x\,dx

    . . . . . . . . I \;=\;\text{-}\csc^3\!x\cot x - 3I + 3\!\int\!\csc^3\!x\,dx

    . . . . . . . 4I \;=\;\text{-}\csc^3\!x\cot x + 3\!\int\!\csc^3\!x\,dx


    Since we know the value of \int\!\csc^3\!x\,dx, we can finish the problem . . .
    That helped a lot! Thank you! But, I think you messed up with distributing your minus sign here:

    Quote Originally Posted by Soroban View Post

    We have: . I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx

    [/size]
    I think it should be a plus and not a minus.
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  5. #5
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    Re: Integration By Parts

    Never mind! ^^
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