Question

Integrate $\int \sec x.\log (\sec x+\tan x)dx$

Answer 1

$I=\int \sec xlog(\sec x+\tan x)dx$ …………….$\left(1\right)$

Using LIATE rule

$I=log(\sec x+\tan x)\int \sec xdx-\int \left[\frac{d}{dx}\right(log(\sec x+\tan x)\int \sec xdx]dx$

$I=\log (\sec x+\tan x)\cdot log(\sec x+\tan x)-\int \left(\frac{\sec x\tan x+{\sec }^{2}x}{\sec x+\tan x}log\right(\sec x+\tan x\left)\right)dx$

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

$I=\left(log\right(\sec x+\tan x){)}^2-I$ …$\{$ From $\left(1\right)$ and $2$ $\}$

$2I=\left[log\right(\sec x+\tan x){]}^2$.