\(\int \frac{1+\sin (\log x)}{1+\cos (\log x)} \mathrm{d} x=\)

\(\begin{aligned}
& \mathrm{I}=\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x \\
& \text { Put } \log x=\mathrm{t} \\
& \Rightarrow x=\mathrm{e}^{\mathrm{t}} \\
& \therefore \quad \mathrm{~d} x=\mathrm{e}^{\mathrm{t}} \mathrm{dt} \\
& \therefore \quad I=\int\left(\frac{1+\sin t}{1+\cos t}\right) \cdot e^t d t \\
& I=\int\left(\frac{\sin ^2 \frac{t}{2}+\cos ^2 \frac{t}{2}+2 \sin \frac{t}{2} \cos \frac{\mathrm{t}}{2}}{2 \cos ^2 \frac{\mathrm{t}}{2}}\right) \mathrm{e}^{\mathrm{t}} \mathrm{dt} \\
& =\frac{1}{2} \int\left(\tan ^2 \frac{\mathrm{t}}{2}+1+2 \tan \frac{\mathrm{t}}{2}\right) \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt} \\
& =\frac{1}{2} \int\left(\sec ^2 \frac{t}{2}+2 \tan \frac{t}{2}\right) e^t d t \\
& =\frac{1}{2} \times 2 \tan \frac{\mathrm{t}}{2} \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{C} \\
& \cdots\left[\int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \cdot \mathrm{f}(x)+\mathrm{c}\right] . \\
& =\tan \frac{t}{2} \cdot e^t+c
\end{aligned}\)
\(I=\tan \left(\frac{\log x}{2}\right) \cdot x+c\)