Question

$\int e^x\left(\tan x-\log \cos x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left(\tan x-\log \cos x\right)dx \\ \mathrm{here}f\left(x\right)=-\log \cos x\mathrm{Put}{e}^{x}f\left(x\right)=t \\ \Rightarrow f\text{'}\left(x\right)=\tan x \\ \mathrm{let}-e^x\log \cos x=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}x \\ -\left[e^x\log \left(\cos x\right)+e^x\frac{1}{\cos x}\times \left(-\sin x\right)\right]=\frac{dt}{dx} \\ \Rightarrow \left[-e^x\log \left(\cos x\right)+e^x\tan x\right]dx=dt \\ \therefore \int e^x\left(\tan x-\log \cos x\right)dx=\int dt \\ =t+C \\ =-e^x\log \left(\cos x\right)+C \\ =e^x\log \left(\sec x\right)+C \end{gathered}$$