We have: $\displaystyle

- \ln \left[ {\sin \left( x \right)} \right] - \ln \left( 2 \right) = \tfrac{{\cos \left( {2x} \right)}}

{1} + \tfrac{{\cos \left( {4x} \right)}}

{2} + \tfrac{{\cos \left( {6x} \right)}}

{3} + ...

$ for $\displaystyle

0 < x < \pi

$

Thus: $\displaystyle

- x \cdot \ln \left[ {\sin \left( x \right)} \right] - x \cdot \ln \left( 2 \right) = \tfrac{{x \cdot \cos\left( {2x} \right)}}

{1} + \tfrac{{x \cdot \cos \left( {4x} \right)}}

{2} + ...

$

Integrate (assuming it's possible to exchange the series and the integral): $\displaystyle

- \int_0^{\tfrac{\pi }

{2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} - \tfrac{{\pi ^2 }}

{8} \cdot \ln \left( 2 \right) = \tfrac{{\int_0^{\tfrac{\pi }

{2}} {x \cdot \cos\left( {2x} \right)dx} }}

{1} + \tfrac{{\int_0^{\tfrac{\pi }

{2}} {x \cdot \cos \left( {4x} \right)dx} }}

{2} + ...

$

Now $\displaystyle

\int_0^{\tfrac{\pi }

{2}} {x \cdot \cos \left( {2 \cdot k \cdot x} \right)dx} = - \left[ {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }}

{{\left( {2k} \right)^2 }}} \right]

$ for all positive integers $\displaystyle k$

Thus: $\displaystyle

\int_0^{\tfrac{\pi }

{2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1}

{4} \cdot \sum\limits_{k = 1}^\infty {\tfrac{{1 + \left( { - 1} \right)^{k + 1} }}

{{k^3 }}} - \tfrac{{\pi ^2 }}

{8} \cdot \ln \left( 2 \right)

$

And: $\displaystyle

\int_0^{\tfrac{\pi }

{2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{1}

{2} \cdot \sum\limits_{k = 1}^\infty {\tfrac{1}

{{\left( {2k + 1} \right)^3 }}} - \tfrac{{\pi ^2 }}

{8} \cdot \ln \left( 2 \right)

$

Since: $\displaystyle

\sum\limits_{k = 1}^\infty {\tfrac{1}

{{\left( {2k + 1} \right)^3 }}} = \tfrac{7}

{8} \cdot \zeta \left( 3 \right)

$ we get $\displaystyle

\int_0^{\tfrac{\pi }

{2}} {x \cdot \ln \left[ {\sin \left( x \right)} \right]dx} = \tfrac{7}

{{16}} \cdot \zeta \left( 3 \right) - \tfrac{{\pi ^2 }}

{8} \cdot \ln \left( 2 \right)

$