Question

Evaluate $\int {{e^x}\sec x\left( {1 + \tan x} \right)dx}$

Answer 1

We will simplify the expression by multiplying $\sec x$ and inside the brackets to form $\int {{e^x}\left( {\sec x + \sec x\tan x} \right)dx}$, which is of the form, $\int {{e^{kx}}\left( {kf\left( x \right) + f'\left( x \right)} \right)dx}$. Now, the value of $\int {{e^{kx}}\left( {kf\left( x \right) + f'\left( x \right)} \right)dx}$ is ${e^{kx}}f\left( x \right) + c$. Hence, substitute the corresponding values to get the required answer.

Complete step-by-step answer:
We have to find the value of $\int {{e^x}\sec x\left( {1 + \tan x} \right)dx}$
First of all, multiply $\sec x$ and inside the brackets to get,
$\int {{e^x}\left( {\sec x + \sec x\tan x} \right)dx}$
Now, we know that $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
This implies we have the integral of the form,
$\int {{e^{kx}}\left( {kf\left( x \right) + f'\left( x \right)} \right)dx}$, where $k = 1$ and $f\left( x \right) = \sec x$
Also, the value of the integral $\int {{e^{kx}}\left( {kf\left( x \right) + f'\left( x \right)} \right)dx}$ is equal to ${e^{kx}}f\left( x \right) + c$, where $c$ is an arbitrary constant.
Hence, the value of $\int {{e^x}\left( {\sec x + \sec x\tan x} \right)dx}$ is ${e^x}\sec x + c$

Note: Do not apply bi-part directly in this question as it will get difficult to solve. Convert the given expression in the form of the known known formula, that is, $\int {{e^{kx}}\left( {kf\left( x \right) + f'\left( x \right)} \right)dx} = {e^{kx}}f\left( x \right) + c$. One must know the formulas of differentiation and integration to do these types of questions correctly.