Question

$\int \left\{\tan \left(\log x\right)+{\sec }^2\left(\log x\right)\right\}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left[\tan \left(\log x\right)+{\sec }^2\left(\log x\right)\right]dx \\ \mathrm{Put}\log x=t \\ \Rightarrow x=e^t \\ \Rightarrow dx=e^{t}dt \\ \therefore I=\int \left(\tan t+{\sec }^{2}t\right)e^{t}dt \\ \mathrm{Here},f\left(t\right)=\tan t \\ \Rightarrow f\text{'}\left(t\right)={\sec }^{2}t \\ \mathrm{let}{e}^t\tan \left(t\right)=p \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}t \\ {e}^t\left[\tan t+{\sec }^{2}t\right]=\frac{dp}{dt} \\ \Rightarrow e^t\left[\tan t+{\sec }^{2}t\right]dt=dp \\ \therefore I=\int dp \\ =p+C \\ =e^t\tan t+C \\ =x\tan \left(\log x\right)+C \end{gathered}$$