Let $ x = \frac {\pi}{4} - \theta \Longrightarrow dx = - \theta ,\ 2x = \frac {\pi}{2} - 2\theta$, $ I = \int_{\frac {\pi}{4}}^0 (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ln \left [1 + \tan \left(\frac {\pi}{4} - \theta \right)\right] d( - \theta)$ $ = \int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ln \frac {2}{1 + \tan \theta} d\theta$ $ = (\ln 2)\int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ d\theta - I$ $ 2I = (\ln 2)\int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ d\theta$ Here, we have : $ \sin ^ 6 \theta + \cos ^ 6 \theta = (\sin ^ 2 \theta + \cos ^ 2 \theta)^3 - 3\sin ^ 2 \theta \cos ^ 2 \theta (\sin ^ 2 \theta + \cos ^ 2 \theta)$ $ = 1 - 3\sin ^ 2 \theta \cos ^ 2 \theta$ $ = 1 - \frac {3}{4}\sin ^ 2 2\theta$ $ = 1 - \frac {3}{4}\cdot \frac {1 - \cos 4\theta}{2}$ $ = \frac {1}{8}(5 + 3\cos 4\theta)$ $ \therefore 2I = \frac {\ln 2}{8}\int_0^{\frac {\pi}{4}} (5 + 3\cos 8\theta)\ d\theta = \frac {5}{32}\pi \ln 2$, yielding $ I = \frac {5}{64}\pi \ln2$.