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Math Help - arctan/arcsin and integration by substitution

  1. #1
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    arctan/arcsin and integration by substitution

    \int \frac{1}{\sqrt{81-x^2}}~dx

    I know I need to use substitution and I am aware of the fact that \frac{1}{\sqrt{1-x^2}} = derivative of arcsin(x), but I am unsure how I apply these to the above problem.

    I have the answer arcsin \left(\frac{x}{9}\right)+C but I don't follow the steps to get to it.
    Last edited by terrorsquid; October 21st 2011 at 12:10 AM.
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  2. #2
    Super Member Quacky's Avatar
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    Re: arctan/arcsin and integration by substitution

    Let I=\int\frac{1}{\sqrt{81-x^2}}dx

    We can do it by substitution, but I wouldn't personally. However, it's good to start from there.

    So I'm going to let x=9sin(u)

    This means that \frac{dx}{du}=9cos(u) so dx=9cos(u)~du

    I=\int\frac{9cos(u)}{\sqrt{81-[9sin(u)]^2}}du

    =\int\frac{9cos(u)}{\sqrt{81-81sin^2(u)}}du

    Factoring the 81:

    =\int\frac{9cos(u)}{\sqrt{81[1-sin^2(u)]}}du

    Applying \sqrt{a\cdot{b}}=\sqrt{a}\sqrt{b} for a,b\geq{0}:

    I=\int\frac{9cos(u)}{9\sqrt{1-sin^2(u)}}du

    But as sin^2(u)+cos^2(u)=1,

    1-sin^2(u)=cos^2(u)

    So I=\int\frac{cos(u)}{\sqrt{cos^2(u)}}du

    =\int\frac{cos(u)}{cos(u)}du

    =\int~du

    =u+C

    Now, I let x=9sin(u) right at the start of my substitution. This means that \frac{x}{9}=sin(u), so u=arcsin(\frac{x}{9})

    Giving us the required final answer of I=arcsin(\frac{x}{9})+C

    Is this clear? Usually I'd ask you to do part of the problem yourself but I don't think there's really a good place to stop with this question.
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  3. #3
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    Re: arctan/arcsin and integration by substitution

    hi,
    the problem is easy
    let y=x/9 => x = 9y and dx = 9dy
    \int \frac{9dy}{\sqrt{81-81y^{2}}}
    \int \frac{9dy}{9\sqrt{1-y^{2}}}
    \int \frac{dy}{\sqrt{1-y^{2}}}
    then
    the solution is : arcsin(y) + c
    i.e. arcsin(x/9) + c
    Last edited by salim; October 21st 2011 at 02:26 AM.
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