Question

Integrate $\int e^{\log (\sec x+\tan x)}\sqrt{1+{\tan }^{2}x}dx$.

Answer 1

We have,

$I=\int e^{\log (\sec x+\tan x)}\sqrt{1+{\tan }^{2}x}dx$

$I=\int e^{\log (\sec x+\tan x)}\sqrt{{\sec }^{2}x}dx$

$I=\int e^{\log (\sec x+\tan x)}\sec xdx$

Let

$t=\log (\sec x+\tan x)$

$\frac{dt}{dx}=\frac{1}{\sec x+\tan x}\times (\sec x\tan x+{\sec }^{2}x)$

$\frac{dt}{dx}=\frac{\sec x}{\sec x+\tan x}\times (\tan x+\sec x)$

$dt=\sec xdx$

Therefore,

$I=\int e^{t}dt$

$I=e^t+C$

On putting the value of $t$, we get

$I=e^{\log (\sec x+\tan x)}+C$

Hence, this is the answer.