Evaluating integral with $\ln x$ Evaluate the following integral:
 $$\int^{\ } \frac{x\ln (x)} {\sqrt{1-x^2}} dx$$
Normally I would show my work so far, however, I don't even know where to start with this question.
 A: You could start with integration by parts; taking $u(x)=\ln x$ and $v'(x) = \frac{x}{\sqrt{1-x^2}}$, with $u'(x) = \frac{1}{x}$ and $v(x) = -\sqrt{1-x^2}$ gives:
$$\int \frac{x \ln x}{\sqrt{1-x^2}} \,\mbox{d}x = -\ln x \sqrt{1-x^2} + \int \frac{\sqrt{1-x^2}}{x}\,\mbox{d}x$$
You can follow up with e.g. $x = \sin t$.
Can you take it from here?
A: This expression heavily hints toward integration by parts. $\ln$ gets differentiated (which makes it much nicer) while the rest is trivially integrated because $x\,dx$ is proportional to the derivative od the thing under the root. In essence,
$$u = \ln x\mapsto du=dx/x$$
$$dv=\frac{x\,dx}{\sqrt{1-x^2}}\mapsto v=-\sqrt{1-x^2}$$
No more logarithm. The integral is still not solved, but you should be able to proceed now.
If you recognize the trigonometric relation $\sqrt{1-x^2}$, imagine $x$ as a trig function, say $x=\cos t$, get an integral that looks like $\tan t \, dt$, which you can recall from a table.
A: $$\int \sin t \log(\sin t)\,dt = \int 2\sin\frac{t}{2}\cos\frac{t}{2}\log\left(2\sin\frac{t}{2}\cos\frac{t}{2}\right)\,dt$$
hence we just need to find a primitive for $z\log z$, that is:
$$ \int z\log z\,dz = -\frac{z^2}{4}+\frac{z^2}{2}\log z $$
by integration by parts.
A: It helps to substitute $x = \sin u$. This gives $\int \sin u \ln \sin u\,du = \frac 1 2 \int \sin u \ln( 1 - \cos^2 u)\,du$. Next, substitute $\cos u = t$ and use that $\ln(1-t^2) = \ln(1+t) + \ln(1-t)$.
I am sure you can finish the calculation.
NB: this method is arguably a bit slower than substituting $t = \sqrt{1-x^2}$ immediately.
A: An alternative simple way is to substitute $u=\sqrt{1-x^2}$ and use $\ln(1-u^2)=\ln(1+u)+\ln(1-u)$.
This gives $$-\frac{1}{2}\int(\ln(1+u)+\ln(1-u))\,\mathrm du$$
which is easy to compute.
