Question

How do you integrate \[{{e}^{\tan x}}.{{\sec }^{2}}\left( x \right).dx\]?

Answer 1

Hint: For solving this question you should know about the integration using by parts. First, in this problem we will assume any term as u and then we put this at that place and differentiate this u and then what there will appear as du then that will be the rest term. Now, make it as a u.du form and solve it and at last put values at their place of u.

Complete step-by-step solution:
According to the question we have to ask to integrate \[{{e}^{\tan x}}.{{\sec }^{2}}\left( x \right).dx\].
So, as we know that for solving this we will use integration by parts. And integration by parts states that we have to make any variable term u in that integration. And then we will differentiate to the u and it will become as du. Then we will change our question in terms of u.du.
Since, it is given that the integral to be evaluate is \[\int{{{e}^{\tan x}}.{{\sec }^{2}}\left( x \right).dx}-\left( 1 \right)\]
Now, if we consider \[\tan x\]as u, because \[{{\sec }^{2}}x\] is the derivative of \[\tan x\],
Thus: \[u=\tan x\]
Now, if we differentiate it,
\[\begin{align}
  & \dfrac{du}{dx}={{\sec }^{2}}x \\
 & du={{\sec }^{2}}x.dx \\
\end{align}\]
So, if we put these value of u and du in equation (1),
\[=\int{{{e}^{u}}.du}\]
If we solve this: \[\int{{{e}^{u}}.du}={{e}^{u}}+c\]
So, if we put the value of \[u=\tan x\] here:
\[={{e}^{\tan x}}+c\]
So, the answer is \[{{e}^{\tan x}}+c\].

Note: While solving this question you have to be careful for every single step. Because if any calculation in any step will be wrong then from there our question will be wrong. And during the considering u choose a value whose derivative is available in the given expression.