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Math Help - Integration

  1. #1
    Member great_math's Avatar
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    Integration

    Please help me with this one:

    \int \log \frac{a}{x}\sin^{-1} (x)\ dx<br />
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  2. #2
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    Interesting. I spent over an hour on this one but I haven't been able to finish it. The only method I can think of is integrating by parts

    int (u dv) = u.v - int (v du)

    Knowing that

    int Arcsin x = sqrt(1-x^2) + x.Arcsin(x) and

    int Log(a/x) = x + xLog(a/x)

    so, if this is correct you just have to pick the right values and continue solving the next integral. I haven't given up on this one so I'll keep trying to solve it.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by great_math View Post
    Please help me with this one:

    \int \log \frac{a}{x}\sin^{-1} (x)\ dx<br />
    This cannot be computed nicely

    \int\ln\left(\frac{a}{x}\right)\arcsin(x)dx
    =\int\bigg[\arcsin(x)\ln(a)-\ln(x)\arcsin(x)\bigg]dx

    Now we are only interested in the second integral. First let

    \arcsin(x)=\varphi\Rightarrow{dx=\cos(\varphi)d\va  rphi}

    Giving us

    \int\ln(x)\arcsin(x)dx\overbrace{\mapsto}^{\arcsin  (x)=\varphi}\int\ln(\sin(\varphi))\varphi\cos(\var  phi)d\varphi

    Integration by parts gives that this is equivalent to

    \varphi\bigg[\sin(\varphi)\ln(\sin(\varphi))-\sin(\varphi))-\int\bigg[\sin(\varphi)\ln(\sin(\varphi))-\sin(\varphi)\bigg]d\varphi

    Now for the second integral we only care about the first term...so in that we let
    x=2\arctan(\psi)

    After simplification this gives us

    \int\sin(\varphi)\ln(\sin(\varphi))d\varphi\overbr  ace{\mapsto}^{\varphi=2\arctan(\psi)}\int\frac{2\p  si}{1+\psi^2}\ln\left(\frac{2\psi}{1+\psi^2}\right  )\frac{2}{1+\psi^2}d\psi

    This looks kind of messy but isnt too bad...do all the undone work...there is your integral
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    I think it can be computed nicely,

    \int{\ln (x)\arcsin x\,dx}=\int{(x\ln x-x)'\arcsin x\,dx}, hence the integral equals x(\ln x-1)\arcsin x-\int{\frac{x\ln x-x}{\sqrt{1-x^{2}}}\,dx}. Now one only cares about \int\frac{x\ln x}{\sqrt{1-x^2}}\,dx, hence, do the partial integration and we get,

    \int{\frac{x\ln x}{\sqrt{1-x^{2}}}\,dx}=-\int{\left( \sqrt{1-x^{2}} \right)'\ln x\,dx}=-\left( \sqrt{1-x^{2}}\ln x-\int{\frac{\sqrt{1-x^{2}}}{x}\,dx} \right),

    where the last integral is elementary.
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