Integrals : 2

**dhiab** · November 24th 2009, 06:00 AM

Calculate integral for the function :

$ \begin{array}{l} \ln (\sin (x)) \\ \ln (\cos (x)) \\ \end{array}$

**Jameson** · November 24th 2009, 07:21 AM

Originally Posted by dhiab

Calculate integral for the function :

$ \begin{array}{l} \ln (\sin (x)) \\ \ln (\cos (x)) \\ \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.

**Drexel28** · November 24th 2009, 07:26 AM

Originally Posted by dhiab

Calculate integral for the function :

$ \begin{array}{l} \ln (\sin (x)) \\ \ln (\cos (x)) \\ \end{array}$

Originally Posted by Jameson

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 $\int\frac{\ln(\sin(x))}{\ln(\cos(x))}dx$ ?

**Jameson** · November 24th 2009, 07:32 AM

Originally Posted by Drexel28

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.

**Drexel28** · November 24th 2009, 07:40 AM

Originally Posted by Jameson

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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