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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