Answer
$\displaystyle \ln\left|\frac{2-\sin 2}{1-\sin 1}\right|\approx 1.929$
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Work Step by Step
Find the indefinite integral first,
$\displaystyle \int\frac{1-\cos\theta}{\theta-\sin\theta}d\theta=\left[\begin{array}{l}
u=\theta-\sin\theta\\
du=(1-\cos\theta)d\theta
\end{array}\right]$
$=\displaystyle \int\frac{1}{u}du=\ln|u|$
$=\ln|\theta-\sin\theta|+C$
Now, the definite integral:
$\displaystyle \int_{1}^{2}\frac{1-\cos\theta}{\theta-\sin\theta}d\theta=[\ln|\theta-\sin\theta|]_{1}^{2}$
$=\ln|2-\sin 2|-\ln|1-\sin 1|$
$=\displaystyle \ln\left|\frac{2-\sin 2}{1-\sin 1}\right|\approx 1.929$