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# Suppose $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx={{e}^{x}}f\left( x \right)+c$. Then determine the function $f\left( x \right)$.

Hint: In this question, we will first evaluate the integral $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx$, for that we will split the integral into $\int {{e}^{x}}\tan x\sec xdx+\int {{e}^{x}}\sec xdx$. Then we will not evaluate the value of the integral $\int {{e}^{x}}\tan x\sec xdx$ rather we will evaluate $\int {{e}^{x}}\sec xdx$ and we will see that it can be expressed in the form of $\int {{e}^{x}}\tan x\sec xdx$. We will then add both the value of the integrals and write it in the form of ${{e}^{x}}f\left( x \right)+c$. Then by equation both the values we will determine the function $f\left( x \right)$.

We are given that $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx={{e}^{x}}f\left( x \right)+c$.
Let $I$ denote the integral $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx$.
That is, let $I=\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx$.
Now on splitting the above integrals, we will have
$I=\int {{e}^{x}}\tan x\sec xdx+\int {{e}^{x}}\sec xdx$
Let us suppose that the integral $\int {{e}^{x}}\tan x\sec xdx$ is denoted by ${{I}_{1}}$ and the integral $\int {{e}^{x}}\sec xdx$ is denoted by ${{I}_{2}}$.
That is, we have
${{I}_{1}}=\int {{e}^{x}}\tan x\sec xdx$ and
${{I}_{2}}=\int {{e}^{x}}\sec xdx$
Since we know that by integration by parts we have $\int{\left( uv \right)dx=u\int{vdx-\int{\dfrac{d}{dx}\left( u \right)\int{vdx}}}}$
We will now evaluate the integral ${{I}_{2}}=\int {{e}^{x}}\sec xdx$ by using integration by parts.
Suppose $u=\sec x$ and $v={{e}^{x}}$, then we have
\begin{align} & {{I}_{2}}=\int {{e}^{x}}\sec xdx \\ & =\sec x\int{{{e}^{x}}dx-\int{\dfrac{d}{dx}\left( \sec x \right)\int{{{e}^{x}}dx}}} \end{align}
Now since $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$, therefore the above integral becomes

\begin{align} & {{I}_{2}}=\sec x\int{{{e}^{x}}dx-\int{\dfrac{d}{dx}\left( \sec x \right)\int{{{e}^{x}}dx}}} \\ & =\sec x\left( {{e}^{x}} \right)-\int{\sec x\tan x{{e}^{x}}dx} \\ & ={{e}^{x}}\sec x-\int{{{e}^{x}}\sec x\tan x{{e}^{x}}dx}+c \end{align}
We also have the integral ${{I}_{1}}=\int {{e}^{x}}\tan x\sec xdx$, hence using the value of ${{I}_{1}}$ in the above integral, we have
\begin{align} & {{I}_{2}}={{e}^{x}}\sec x-\int{{{e}^{x}}\sec x\tan x{{e}^{x}}dx}+c \\ & ={{e}^{x}}\sec x-{{I}_{1}}+c \end{align}
Now substituting the value of integral ${{I}_{1}}$ and ${{I}_{2}}$ in integral $I$, we get
\begin{align} & I=\int {{e}^{x}}\tan x\sec xdx+\int {{e}^{x}}\sec xdx \\ & ={{I}_{1}}+{{I}_{2}} \\ & ={{I}_{1}}+{{e}^{x}}\sec x-{{I}_{1}}+c \\ & ={{e}^{x}}\sec x+c..............(1) \end{align}
We are also given that $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx={{e}^{x}}f\left( x \right)+c$.
That is $I={{e}^{x}}f\left( x \right)+c.............(2)$.
On equating equation (1) and (2), we get
${{e}^{x}}f\left( x \right)+c={{e}^{x}}\sec x+c$
$\Rightarrow f\left( x \right)=\sec x$
Hence we get that the function $f\left( x \right)$ such that $\int {{e}^{x}}\left( \tan x+1 \right)\sec xdx={{e}^{x}}f\left( x \right)+c$ is equals to $\sec x$.

Note:
In this problem, while evaluating the integrals $I=\int {{e}^{x}}\tan x\sec x dx+\int {{e}^{x}}\sec xdx$ where ${{I}_{1}}=\int {{e}^{x}}\tan x\sec xdx$ and ${{I}_{2}}=\int {{e}^{x}}\sec x dx$, please do not try to expand the integral ${{I}_{1}}=\int {{e}^{x}}\tan x\sec xdx$, otherwise complications of the problem will increase. Moreover keep in mind the fact that $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$ and use it in order to simplify the evaluation of the integrals.