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Math Help - Integrals : 2

  1. #1
    Super Member dhiab's Avatar
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    Integrals : 2

    Calculate integral for the function :

    <br />
\begin{array}{l}<br />
\ln (\sin (x)) \\ <br />
\ln (\cos (x)) \\ <br />
\end{array}
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  2. #2
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    Quote Originally Posted by dhiab View Post
    Calculate integral for the function :

    <br />
\begin{array}{l}<br />
\ln (\sin (x)) \\ <br />
\ln (\cos (x)) \\ <br />
\end{array}
    Mathematica got a nasty result for \int \ln(\sin(x))dx involving i and non-elementary functions, so I don't know if I am misunderstanding what you are asked to do. If you meant a fraction, Mathematica can't solve it.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dhiab View Post
    Calculate integral for the function :

    <br />
\begin{array}{l}<br />
\ln (\sin (x)) \\ <br />
\ln (\cos (x)) \\ <br />
\end{array}
    Quote Originally Posted by Jameson View Post
    Mathematica got a nasty result for \int \ln(\sin(x))dx involving i and non-elementary functions, so I don't know if I am misunderstanding what you are asked to do. If you meant a fraction, Mathematica can't solve it.
    Did the OP perhaps mean \int\frac{\ln(\sin(x))}{\ln(\cos(x))}dx?
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Did the OP perhaps mean \int\frac{\ln(\sin(x))}{\ln(\cos(x))}dx?
    I plugged that in as well and Mathematica couldn't find a solution. I haven't tried to work it out myself.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jameson View Post
    I plugged that in as well and Mathematica couldn't find a solution. I haven't tried to work it out myself.
    Maybe we can make that a tad more apparent? Let x=\arcsin(z)\implies dx=\frac{dz}{\sqrt{1-z^2}} so our integral becomes \int\frac{\ln(z)}{\sqrt{1-z^2}\ln\left(\sqrt{1-z^2}\right)}. Let \sqrt{1-z^2}=\tau\implies \frac{-\tau}{\sqrt{1-\tau^2}}d\tau=dz. So then our integral becomes \frac{-1}{2}\int\frac{\ln\left(1-\tau^2\right)}{\sqrt{1-\tau^2}\ln\left(\tau\right)}d\tau. That looks almost doable. I am not feeling it right now. Maybe the OP can continue.
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