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Math Help - Integration by parts

  1. #1
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    Integration by parts

    I'm trying to integrate this function using integration by parts, but I get stuck after a while.

    The function is

    \int sin^2(\theta) d\theta

    I start by using the formula \int f g' dx = fg-\int g f'dx

    So taking f = g' = sin(\theta), then f' = cos(\theta), g = -cos(\theta)

    \int f g' dx = fg-\int g f'dx

     = -cos(\theta)sin(\theta) -\int -cos(\theta)cos(\theta) d\theta

    Now what do I do?
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  2. #2
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    No need for integration by parts, use trigonometric identities.

    \int{\sin^2{\theta}\,d\theta} = \int{\frac{1}{2} - \frac{1}{2}\cos{2\theta}\,d\theta}

     = \frac{1}{2}\theta - \frac{1}{4}\sin{2\theta} + C.
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  3. #3
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    The question states to use integration by parts, I assume there would be a way?
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  4. #4
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    You may need to integrate by parts twice here.

    \displaystyle \int \sin^2 \theta~d\theta = \int \sin \theta\times \sin \theta~d\theta

    \displaystyle   du = \sin \theta\implies u = -\cos\theta and \displaystyle   v = \sin \theta\implies dv = \cos\theta

    Now \displaystyle \int du \times v = uv -\int dv \times u

    What do you get?
    Last edited by pickslides; August 19th 2010 at 06:17 PM.
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  5. #5
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    Quote Originally Posted by exothesis View Post
    I'm trying to integrate this function using integration by parts, but I get stuck after a while.

    The function is

    \int sin^2(\theta) d\theta

    I start by using the formula \int f g' dx = fg-\int g f'dx

    So taking f = g' = sin(\theta), then f' = cos(\theta), g = -cos(\theta)

    \int f g' dx = fg-\int g f'dx

     = -cos(\theta)sin(\theta) -\int -cos(\theta)cos(\theta) d\theta

    Now what do I do?
    So you have \int sin^2(\theta) d\theta= -cos(\theta)sin(\theta)+ \int cos^2(\theta) d\theta

    Now do it again: integrate \int cos^2(\theta)d\theta) by parts, taking u= cos(\theta), dv= cos(\theta)d\theta.

    You may get a \int sin^2(\theta)d\theta) again but now you can solve the equation for \int sin^2(\theta)d\theta.
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  6. #6
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    I end up with 0? I don't think I'm doing it right. This is what I do:
    \int cos^2(\theta)d\theta = cos(\theta)sin(\theta)+ \int sin^2(\theta) d\theta

    But then all together that means:
     \int sin^2(\theta) d\theta= -cos(\theta)sin(\theta)+ cos(\theta)sin(\theta)+ \int sin^2(\theta) d\theta = 0

    What am I doing wrong here? My CAS calculator says the answer should be
    \frac{x}{2} - \frac{sin\theta cos\theta}{2}
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  7. #7
    Junior Member BayernMunich's Avatar
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    Solving this one with integration by parts is a time's wasting.

    Use this:

    sin^2(\theta)=\dfrac{1-cos(2\theta)}{2}
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  8. #8
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    Quote Originally Posted by BayernMunich View Post
    Solving this one with integration by parts is a time's wasting.

    Use this:

    sin^2(\theta)=\dfrac{1-cos(2\theta)}{2}

    Quote Originally Posted by exothesis
    The question states to use integration by parts, I assume there would be a way?
    .
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  9. #9
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    http://www.ballooncalculus.org/forum...lllpartslllone

    Ok, it's a video, but my link is to a .wmv file and I need to do it differently, clearly. Will fix. (Edit: fixed, but you have to follow the link to see the video - was hoping it would project right here!)

    Anyway, just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards or integrate up with respect to the main variable (in this case theta).

    The general drift is...



    ________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; August 26th 2010 at 04:37 AM.
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  10. #10
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    If you MUST use integration by parts, then rewrite your integral as

    \int{\sin^2{\theta}\,d\theta} = \int{1 - \cos^2{\theta}\,d\theta}

     = \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta}


    Now apply integration by parts with u = 1 - \cos{\theta} so du = \sin{\theta} and dv = 1 + \cos{\theta} so v = \theta + \sin{\theta}.

    The integral becomes

    \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta} = (1 - \cos{\theta})(\theta + \sin{\theta}) - \int{\sin{\theta}(\theta + \sin{\theta})\,d\theta}

     = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}



    So this gives

    \int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}

    2\int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}

    \int{\sin^2{\theta}\,d\theta} = \frac{1}{2}\left(\theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}\right).


    Apply integration by parts once more, then you will have your answer.
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  11. #11
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    Quote Originally Posted by Prove It View Post
    If you MUST use integration by parts, then rewrite your integral as

    \int{\sin^2{\theta}\,d\theta} = \int{1 - \cos^2{\theta}\,d\theta}

     = \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta}


    Now apply integration by parts with u = 1 - \cos{\theta} so du = \sin{\theta} and dv = 1 + \cos{\theta} so v = \theta + \sin{\theta}.

    The integral becomes

    \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta} = (1 - \cos{\theta})(\theta + \sin{\theta}) - \int{\sin{\theta}(\theta + \sin{\theta})\,d\theta}

     = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}



    So this gives

    \int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}

    2\int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}

    \int{\sin^2{\theta}\,d\theta} = \frac{1}{2}\left(\theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}\right).


    Apply integration by parts once more, then you will have your answer.
    A scenic route?!

    Surely, just convert cos squared to 1 minus sin squared as soon as you encounter it in the original version of parts. Sorry my web host have a problem but the pics will hopefully re-appear in a short while.
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  12. #12
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    Thanks for all the help. I should have no trouble solving it using the methods described above.
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  13. #13
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    Hello, exothesis!

    The formula I use has this form: . \int u\,dv \;=\;uv - \int v\,du



    Integrate by parts: . I \;=\;\int \sin^2x \, dx

    By parts: . \begin{array}{cccccccc}<br />
u &=& \sin x && dv &=& \sin x\,dx \\<br />
du &=& \cos x\,dx && v &=& -\cos x \end{array}


    And we have: . I \;=\;-\sin x\cos x - \int(-\cos x)\cos x\,dx

    . . . . . . . . . . I \;=\;-\sin x\cos x + \int\cos^2x\,dx

    . . . . . . . . . . I \;=\;-\sin x\cos x + \int(1 - \sin^2x)\,dx

    . . . . . . . . . . I \;=\;-\sin x\cos x + \int dx - \underbrace{\int\sin^2x\,dx}_{\text{This is }I}


    So we have: . I \;=\;-\sin x\cos x + \int dx - I

    . . . . . . . . . 2I \;=\;-\sin x\cos x + \int dx

    . . . . . . . . . 2I \;=\;-\sin x\cos x + x + C


    Therefore: . . I \;=\;\frac{1}{2}(x - \sin x\cos x) + C

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