Seems too easy!

OK Here we begin … Let $\displaystyle 2x = t$ ,then $\displaystyle dx = \frac{1}{2} dt$ ,and our integral changes as $I = \frac{1}{8} \int_{0}^{^\pi/_2} \dfrac{t^2 (\sin t - \cos t)}{(1 + \sin t)(\cos^2 (^t/_2))} dx$ $I = \dfrac{1}{4} \int_{0}^{^\pi/_2} \dfrac{t^2 (\sin t - \cos t)}{(1 + \sin t)(1 + \cos t)} dx………….(1)$ Now using the properties of definite integral we get $I = \dfrac{1}{4} \int_{0}^{^\pi/_2} \dfrac{\Bigg( \dfrac{\pi}{2} - t\Bigg)^{2} (\cos t - \sin t)}{(1 + \cos t)(1 + \sin t)} dx…………….....(2)$ Adding (1)&(2) we get $2I = \dfrac{1}{4} \int_{0}^{^\pi/_2} \dfrac{\Bigg( \pi t - \dfrac{\pi^{2}}{4}\Bigg) (\sin t - \cos t)}{(1 + \sin t)(1 + \cos t)} dx$ But $\int_{0}^{^\pi/_{2}} \dfrac{\sin t - \cos t}{(1 + \sin t)(1 + \cos t) } dx = 0$ And hence after distributing we get $I = \dfrac{1}{8} \int_{0}^{^\pi/_2} \dfrac{\pi t (\sin t - \cos t)}{(1 + \sin t)(1 + \cos t) } dx$ Hence after solving this we get $I = \boxed{ \dfrac{\pi^2}{16} - \dfrac{\pi}{4} \log2 }$