Showing $\ln(\sin(x))$ is in $L_1$ 
Prove $\ln[\sin(x)] \in L_1 [0,1].$

Since the problem does not require actually solving for the value, my strategy is to bound the integral somehow. I thought I was out of this one free since for $\epsilon > 0$ small enough, $$\lim_{\epsilon \to 0}\int_\epsilon^1 e^{\left|\ln(\sin(x))\right|}dx=\cos(\epsilon)-\cos(1) \to 1-\cos(1)<\infty$$
and so by Jensen's Inequality, $$e^{\int_0^1 \left| \ln(\sin(x))\right|\,dx}\le \int_0^1e^{\left|\ln(\sin(x))\right|}\,dx\le1-\cos(1)<\infty$$ so that $\int_0^1 \left|\ln(\sin(x))\right|\,dx<\infty$. 
The problem, of course, is that the argument begs the question, since Jensen's assumes the function in question is integrable to begin with, and that's what I'm trying to show. 
Any way to save my proof, or do I have to use a different method? I attempted integration by parts to no avail, so I am assuming there is some "trick" calculation I do not know that I should use here. 

Here is a proof that hides behind a theorem on swapping order of integration:
We have $0 \le {x \over 2} \le \sin x$, and $-\log$ is decreasing on $(0,1]$.
Then $\int_0^1 | \log(\sin x)| dx = \int_0^1 - \log( \sin x) dx \le \int_0^1 - \log( { x \over 2}) dx  = \log 2 + \int_0^1 - \log( { x}) dx$.
Tonelli gives $\int_0^1 -\log(x) dx = \int_{x=0}^1 \int_{t=x}^1 {dt \over t} dx = \int_{t=0}^1 \int_{x=0}^t {dx \over t} dt = 1$.
Hence the upper bound $\int_0^1 | \log(\sin x)| dx \le 1+ \log 2$.
