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

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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 \[f\left( x \right)\].

Complete step by step answer:
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
  & {{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}}}
Now since \[\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x\], therefore the above integral becomes

  & {{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
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
  & {{I}_{2}}={{e}^{x}}\sec x-\int{{{e}^{x}}\sec x\tan x{{e}^{x}}dx}+c \\
 & ={{e}^{x}}\sec x-{{I}_{1}}+c
Now substituting the value of integral \[{{I}_{1}}\] and \[{{I}_{2}}\] in integral \[I\], we get
  & 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)
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 \[\sec x\].

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.