This integral has everything

I have uploaded this picture file, because the LATEX tag doesn't work for me at the moment.

$\text {let }I=\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\ln { \left( \frac { \tan t }{ e^{ t } } \right) } dt } \\ I=\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\ln { \left( \frac { \cot t }{ e^{ \pi /2-t } } \right) } dt } \qquad \qquad \color{#3D99F6} //put \quad u=\pi /2-t \\ 2I=\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\ln { \left( \frac { \tan t }{ e^{ t } } \right) } dt } +\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\ln { \left( \frac { \cot t }{ e^{ \pi /2-t } } \right) } dt } \\ 2I=\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\left( \ln { \left( \frac { \tan t }{ e^{ t } } \right) } +\ln { \left( \frac { \cot t }{ e^{ \pi /2-t } } \right) } \right) dt } =\int _{ 0 }^{ \pi /2 }{ \sin t\cos t\ln { \left( e^{ -\pi /2 } \right) } dt } \\ 2I=-\frac { \pi }{ 2 } \int _{ 0 }^{ \pi /2 }{ \sin t\cos t\quad dt } =-\frac { \pi }{ 4 } \\ I=-\frac { \pi }{ 8 }$

Another trick is to expand the integrand into:

sin(x) cos(x) ln[ sin(x)/(cos(x) e^x) ] = sin(x) cos(x)*[ln sin(x) - ln cos(x) - ln e^x];

or sin(x) cos(x) ln sin(x) - sin(x) cos(x) ln cos(x) - sin(x) cos(x) x;

or sin(x) cos(x) ln sin(x) - sin(x) cos(x) ln cos(x) - (x/2)*sin(2x); (i).

Now, taking u = sin(x); du = cos(x) dx; v = cos(x); dv = -sin(x) dx, the integral of u*ln(u) becomes:

(u^2)/2 * [ln(u) - 1/2] (ii)

using integration-by-parts. This yields the final integration value of:

(1/2) (sin(x))^2 * [ln(sin(x) - 1/2] + (1/2) (cos(x))^2 * [ln(cos(x) - 1/2] - (1/8) sin(2x) + (x/4) cos(2x) (iii)

Plugging in the bounds 0 and pi/2 to (iii) ultimately gives -pi/8 for a result.