∫ π 4 5 π 4 cos t sin t + sin t cos t d t =

$I={\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\left(\left|\cos t\right|\sin t+\left|\sin t\right|\cos t\right)dt$

$={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\cos t\sin tdt-{\int }_{\frac{\mathrm{\pi }}{2}}^{\frac{5\mathrm{\pi }}{4}}\cos t\sin tdt+{\int }_{\frac{\mathrm{\pi }}{4}}^{\mathrm{\pi }}\sin t\cos tdt-{\int }_{\mathrm{\pi }}^{\frac{5\mathrm{\pi }}{4}}\sin t\cos tdt$

$={\left[\frac{t^2}{2}\right]}_{\frac{1}{\sqrt{2}}}^1-{\left[\frac{t^2}{2}\right]}_1^{-\frac{1}{\sqrt{2}}}+{\left[\frac{t^2}{2}\right]}_{\frac{1}{\sqrt{2}}}^0-{\left[\frac{t^2}{2}\right]}_0^{-\frac{1}{\sqrt{2}}}$ (let $\sin t=t$)

$=\frac{1}{2}-\frac{1}{4}-\frac{1}{4}+\frac{1}{2}+0-\frac{1}{4}-\frac{1}{4}+0=0$

$=\frac{5{\mathrm{\pi }}^2}{4}-\frac{13{\mathrm{\pi }}^2}{8}$