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Math Help - Integration by parts with substitution

  1. #1
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    Integration by parts with substitution

    Integration by parts I understand, but part of the problems for this section has instructions to "First make a substitution and then use integration by parts to evaluate the integral". The book itself doesn't even mention this in the section. I have the answers, but it's not really helping me understand what's going on or why.

    example:
    \int{cos\sqrt{x}dx}

    The answer says to substitute for \sqrt{x}, so
    u=x^\frac{1}{2}.
    Thus
    du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}

    Switching in the y is a bit of a stretch. I understand it, but I'm not sure I understand why. The book then shows to substitute all this back into the original problem to get

    \int{cosy(2ydy)}

    I don't understand this at all... how'd we get from  dx = \frac{1}{2y}dy to cosy(2ydy)? From there, it's just integration by parts, and I understand the process. It's only the substitution part that's got me baffled.

    If someone is able to explain the how and why of that, I'd be much appreciative.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Integration by parts with substitution

    If you have:
    \int \cos\left(\sqrt{x}\right)dx
    Let \sqrt{x}=t \Rightarrow \frac{dx}{2\sqrt{x}}=dt \Leftrightarrow dx=2tdt
    So the integral can be written as:
    \int \cos(t)2tdt

    Note:
    If you use the variable u then it's confusing to suddenly use the variable y so stay with one variable.
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  3. #3
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    Re: Integration by parts with substitution

    Quote Originally Posted by satis View Post
    Integration by parts I understand, but part of the problems for this section has instructions to "First make a substitution and then use integration by parts to evaluate the integral". The book itself doesn't even mention this in the section. I have the answers, but it's not really helping me understand what's going on or why.

    example:
    \int{cos\sqrt{x}dx}

    The answer says to substitute for \sqrt{x}, so
    u=x^\frac{1}{2}.
    Thus
    du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}
    This is completely wrong because you forgot the "dx" when you differentiated. You should have
    du= \frac{1}{2}x^{\frac{-1}{2}}dx
    from that, 2x^{\frac{1}{2}}du= dx or 2udu= dx.

    Switching in the y is a bit of a stretch. I understand it, but I'm not sure I understand why. The book then shows to substitute all this back into the original problem to get

    \int{cosy(2ydy)}

    I don't understand this at all... how'd we get from  dx = \frac{1}{2y}dy to cosy(2ydy)? From there, it's just integration by parts, and I understand the process. It's only the substitution part that's got me baffled.

    If someone is able to explain the how and why of that, I'd be much appreciative.
    Follow Math Help Forum on Facebook and Google+

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