\(\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=\)

Let \(\begin{aligned} & I=\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x \\ &=\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) \mathrm{d} \theta \\ &=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] \mathrm{d} \theta \\ & \cdots \cdot\left[\because \int_0^a f(x) \mathrm{d} x=\int_0^a \mathrm{f}(\mathrm{a}-x) \mathrm{d} x\right]\end{aligned}\)
\(\begin{aligned} & =\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta \\ & =\int_0^{\frac{\pi}{4}} \log 2 d \theta-\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta \\ \therefore \quad & 2 I=\int_0^{\frac{\pi}{4}} \log 2 d \theta \Rightarrow I=\frac{\log 2}{2}[\theta]_0^{\pi / 4}=\frac{\pi}{8} \log 2\end{aligned}\)