# Math Help - Integrals : 2

1. ## Integrals : 2

Calculate integral for the function :

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

2. 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.

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

4. 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.

5. 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.