Mathematics

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

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

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

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

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

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

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

$dt=\sec x\ dx$

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.

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Subjective Medium Published on 17th 09, 2020
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