Integrate $\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$. HW Since this is a homework problem, a hint would be appreciated to help me get this started, since I have no idea how to start this. Thanks
Here's the problem:
Compute the improper integral:
$$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$$
 A: Here is the full solution
Using the substitution $x=\sin \theta$, the limit of the integrand changes from $x=0$ to $\theta =0$ and from $x=1$ to $\theta = \frac{\pi}{2}$ since $\theta=\arcsin(x)$.
Also, $dx=\cos \theta d\theta\Leftrightarrow  dx=\sqrt {1 - {{\sin }^2}\theta } d\theta \Leftrightarrow dx=\sqrt {1 - {x^2}} d\theta \Leftrightarrow d\theta=\frac{{dx}}{{\sqrt {1 - {x^2}} }}$.
So we have,
$$I = \int\limits_0^1 {\frac{{\ln x}}{{\sqrt {1 - {x^2}} }}} dx = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x)dx} $$
from the properties of definite integral, recall that
$$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)dx} $$
$$ \Rightarrow I=\int\limits_0^{\frac{\pi }{2}} {\ln (\sin x)dx}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {\frac{\pi }{2} - x} \right))dx}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\cos x)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x) + \ln \left( {\cos x} \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x\cos x)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\frac{{\sin \left( {2x} \right)}}{2})dx}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx}  - \int\limits_0^{\frac{\pi }{2}} {\ln (2)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx}  - \int\limits_0^{\frac{\pi }{2}} {\ln (2)dx}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx}  - \frac{\pi }{2}\ln \left( 2 \right)$$
In the $\int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} $, if we let $2x = t \Rightarrow dx = \frac{{dt}}{2}$ and the limit of the integrand becomes $t = 0$ and $t=\pi$. That is, $\int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx}  = \frac{1}{2}\int\limits_0^\pi  {\ln (\sin \left( t \right))dt}$
Since
$$\int\limits_0^a {f\left( x \right)} dx = 2\int\limits_0^{{\textstyle{a \over 2}}} {f\left( x \right)} dx \quad\mbox{ if } f\left( {a - x} \right) = f\left( x \right)$$
$$ \Rightarrow \frac{1}{2}\int\limits_0^\pi  {\ln (\sin \left( t \right))dt}  = \frac{2}{2}\int\limits_0^{{\textstyle{\pi  \over 2}}} {\ln (\sin \left( t \right))dt}  = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( x \right))dx} \quad \mbox{ since } \sin \left( {\pi  - x} \right) = \sin \left( x \right) \mbox{ and $t$, $x$ are dummy variables}$$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx}  - \frac{\pi }{2}\ln \left( 2 \right) = I - \frac{\pi }{2}\ln \left( 2 \right)$$
$$ \Rightarrow I =  - \frac{\pi }{2}\ln \left( 2 \right) = \frac{\pi }{2}\ln \left( {\frac{1}{2}} \right)$$
$$ \Rightarrow \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( x \right))dx}  = \frac{\pi }{2}\ln \left( {\frac{1}{2}} \right)$$
Any Question?
A: The integral you provide is
$$
I\equiv \int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx.
$$
We can change variables by $x=\sin \phi$, $dx=\cos \phi d \phi$ and use $1-\sin^2 \phi=\cos^2 \phi$.  Your integral becomes 
$$
I=\int_0^{\pi/2} \ln (\sin \phi)\, d\phi 
$$
So now you have to solve this integral, since this is homework, I will give you a small hint.  Consider
$$
2I=\int_0^{\pi/2}\ln (\sin \phi) \, d\phi+\int_0^{\pi/2}\ln (\cos \phi) \, d\phi.
$$
This is true by symmetry of the $\ln \cos \phi$ and $\ln \sin \phi$ function for the region of integration $\phi\in[0,\pi/2]$. You should be able to take it from here by using standard log rules and trig rules.  Let me know if you need more help
