another definite integral (3 may)

\sin^4x + \cos^4 x \begin{align*} & \\ & \\ & = 1-\frac{\sin^2 2x}{2} \\ &= \frac{1+(1-\sin^2 2x)}{2} \\ &= \frac{1+\cos^2 2x}{2} \end{align*}

so \begin{align*} & \mathbb{I} = \int_{0}^{\frac{\pi}{4}} \ln(2) + \int_{0}^{\frac{\pi}4} \ln [\frac{1+\sin^2 2x}{1+\cos^2 2x}] \\ & =\frac{\pi}{4} \ln 2 + \mathbb{I}_1 \\ & \mathbb{I}_1 , \\ & \texttt{take 2x=t}\\ &\mathbb{I}_1 = \int_{0}^{\frac{\pi}{2}} \ln [\frac{1+ \sin ^2 t}{1+ \cos^2 t}] \\ & \texttt{using property replace }t = \frac{\pi}{2}-t\\ & \implies \mathbb{I}_1 = \mathbb{-I}_1 \\ & \implies \mathbb{I}_1 = 0 \\ & .:. \mathbb{I}={\frac{\pi}{4} \ln 2}\\ & \Box \\ & \\ & \\ & \\ & \end{align*}