Definite integral $\int_0^{\pi} \sin^{n} x \ln(\sin x) dx$? Is the integral
$$I(n) = \int_0^{\pi} \sin^{n}(x) \ln(\sin(x)) dx$$
analytically tractable for $n \in \mathbb{N}$? If not, are there good upper and lower bounds?

To clarify, although this has a fairly standard-looking form, I haven't been able to find an answer anywhere and I've tried a few of the usual tricks (although I'm a tad rusty).
I have tried integration by parts to reduce to a simpler or recursive form but this seems to make the expression more complex, and likewise with the substitution $u=\sin(x)$.

Equally, it is relatively easy to see that the expression under the integral is non-positive. Taking a derivative
$$ \frac{d}{dx} \sin^{n}(x) \ln(\sin(x)) = \sin^{n-1}(x) \cos(x) (1 + n \ln(\sin(x))) = 0$$
gives a has minimum at $-\frac{1}{en}$.
Together these give the reasonably simple bounds: $$-\frac{\pi}{en} < I(n) < 0.$$
But numerically these don't appear to be particularly tight, so I'm still hopeful of an exact solution or an improvement.
 A: Integrate by parts to obtain a recursive formula as follows
\begin{align}
I_n&=\int_0^{\pi} \sin^{n}x \ln(\sin x) dx\\
&= -\int_0^{\pi} \sin^{n-1}x \ln(\sin x)\> d(\cos x)\>\\
& =\int_0^{\pi}((n-1) \sin^{n-2}x \cos^2x\ln(\sin x)+ \sin^{n-2}x\cos^2x)dx\\
&= (n-1) (I_{n-2}-I_n)+ \frac1{n-1}\int_0^{\pi}\sin^{n}x\>dx
\end{align}
Thus
$$I_n = \frac{n-1}n I_{n-2} +\frac1{n(n-1)} \int_0^{\pi}\sin^{n}x\>dx
$$
with $I_0 = -\pi\ln2$ and $I_1= \ln2 -1$. (See here for evaluating $\int_0^{\pi/2}\sin^{n}x\>dx$.)
A: As was hinted at in @Quanto's answer, we have the Wallis integral
$$W(\alpha) := \int_0^{\pi/2} \sin^\alpha(x)\ dx = \frac{\sqrt \pi}{2} \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)}, $$
for all $\alpha \geqslant 0$. Differentiating $W$ with respect to $\alpha$, we find (by the symmetry of the integrand)
$$\frac{dW}{d\alpha} (\alpha) = \int_0^{\pi/2} \sin^{\alpha}(x) \log(\sin(x))\ dx = \frac 1 2 I(\alpha), $$
so that, setting $g(\alpha) := \Gamma\left(\dfrac{\alpha+1}{2}\right)\Big/\Gamma\left(\dfrac \alpha 2 + 1\right)$,
$$\begin{split}
\frac{1}{\sqrt{\pi}}I(\alpha) &= \frac{dg}{d\alpha} (\alpha) = \frac{d}{d\alpha}e^{\log g(\alpha)} = e^{\log g(\alpha)} \frac{d}{d\alpha}\log(g(\alpha)) \\
&=g(\alpha) \frac{d}{d\alpha}\left\{ \log \Gamma\left(\frac{\alpha+1}{2}\right) -\log\Gamma\left(\frac \alpha 2 + 1\right)\right\} \\
&= \frac 1 2\frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)} \left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\},
\end{split}$$
where $\psi^{(0)}(z)=(\log \Gamma)'(z)$ is the digamma function.

Bonus. We can go further. Using the identity $\psi^{(0)}(t+1) = \psi^{(0)}(t) + 1/t$ for all $t>0$, and taking advantage of the fact that
$$\psi^{(0)}\left(\frac{s+1}{2s} \right) - \psi^{(0)}\left(\frac{1}{2s} \right) = \sum_{k=0}^\infty \frac{(-1)^k}{sk+1}$$
for $s>0$, choosing $s = (\alpha+2)^{-1}$ we find
$$\frac{\alpha+2}{2}\left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\} = - \frac{\alpha+2}{\alpha+1} + \sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1},  $$
whence
$$\begin{split}
I(\alpha) &=  \sqrt\pi g(\alpha)\left\{-\frac1{\alpha+1} + \frac{1}{\alpha+2}\sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1}\right\}\\
&=\sqrt\pi g(\alpha) \sum_{k=1}^\infty \frac{(-1)^k}{\alpha+k}.
\end{split}$$
Now we restrict ourselves to nonnegative integers $\alpha = n \geqslant 1$, and we observe that
$$\sum_{k=0}^\infty \frac{(-1)^k}{n+k} = \sum_{j=n}^\infty \frac{(-1)^{j-n}}{j} = (-1)^{n+1} \sum_{j=n}^\infty \frac{(-1)^{j+1}}{j} =(-1)^{n+1}\left[-\tilde H_{n} + \log(2) \right], $$
where $\tilde H_{n} = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$ is the $n$-th alternating harmonic number. we obtain
$$\boxed{I(n) =   (-1)^n\sqrt\pi \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac n 2 + 1\right)} \big( \tilde H_n - \log(2)\big).}$$
One may then distinguish even-$n$ and odd-$n$ cases and employ the duplication formula for $\Gamma$ to express $I(n)$ in terms of factorials and double factorials.
A: Mathematica says:
$$\fbox{$\frac{\sqrt{\pi } \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right) \Gamma
   \left(\frac{n+1}{2}\right)}{n \Gamma \left(\frac{n}{2}\right)}\text{ if }\Re(n)>-1$}.$$
A: We have
$$
I_n=2\int_0^{\pi/2}\sin^n(x)\log(\sin(x)) dx=
\int_0^{\pi/2}\sin^n(x)\log(1-\cos^2(x)) dx
$$
$$
=-\sum_{k\geq 1}\frac 1k\int_0^{\pi/2}\sin^n(x)\cos^{2k}(x) dx.
$$
With the help of Euler's Beta function
$$
\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}=B(u,v)=2\int_0^{\pi/2}\sin^{2u-1}(x)\cos^{2v-1}(x) dx
$$
we obtain
$$
I_n=-\Gamma(\frac {n+1}2)\sum_{k\geq 1}\frac 1{2k}\frac{\Gamma(k+\frac 12)}{\Gamma(\frac n2+k+1)}.
$$
This can be further evaluated in closed terms by means of Harmonic Numbers of integer / half integer arguments, as in Igor Rivin's answer, but it may be already good for estimates.
