Evaluate the indefinite integral of $\int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx}$ where functions are well defined. \[\]

Hint: We recall the definition of indefinite integration and the substitution by integration method to solve the indefinite integration. We in order to substitute ${{e}^{x}}=u$ in the integrand of the function find $du$ and then substitute${{e}^{x}}=u$. We use the standard integration $\int{\sec x\tan x}dx=\sec x+c$ to find the required integral. \[\]

Complete step by step answer:
We know that an antiderivative, primitive function or indefinite integral of a function $f$ is a differentiable function $F$ whose derivative is equal to the original function $f$ which means${{F}^{'}}=f$. The process of finding integral is called integration and the original function is called integrand. We write integration in variable $x$ as
\[\int{f\left( x \right)}dx=F\left( x \right)+c\]
 Here $c$ is an arbitrary constant of integration. We also integral remains the same even if we change the variable. It means for any variable $u$ we have;
\[\int{f\left( u \right)}du=\int{f\left( u \right)}du\]
If we have composite function $f\left( g\left( x \right) \right)$and the differential of the function inside the bracket ${{g}^{'}}\left( x \right)$we can substitute the $g\left( x \right)$ as by variable $u$ which means$g\left( x \right)=u$we can integrate as
\[\int{f\left( g\left( x \right) \right){{g}^{'}}\left( x \right)}dx=\int{f\left( u \right)}du\]
The above method is called integration by substitution, u-substitution or change of variable method.
We are given in the question to evaluate the indefinite integral of the following integration$\int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx}$. We see that here the integrand is ${{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}$. We see that the function ${{e}^{x}}$ is repeated 3 times and as well as we know that $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$. So we solve the problem by the u-substitution method. We in order to substitute ${{e}^{x}}=u$ in the integrand of the function find $du$ by differentiating with respect to $u$. We have;
\[\begin{align}
  & \dfrac{d}{dt}{{e}^{x}}=\dfrac{du}{dx} \\
 & \Rightarrow {{e}^{x}}=\dfrac{du}{dx} \\
 & \Rightarrow du={{e}^{x}}dx \\
\end{align}\]
Now we put $u={{e}^{x}},du={{e}^{x}}dx$ in the integrand to have;
\[\begin{align}
  & \Rightarrow \int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx} \\
 & \Rightarrow \int{\tan {{e}^{x}}\sec {{e}^{x}}\times {{e}^{x}}dx} \\
 & \Rightarrow \int{\tan u\sec udu} \\
\end{align}\]
We use the standard integration $\int{\sec x\tan x}dx=\sec x+c$ in the above step to have;
\[\Rightarrow \sec u+c\]
We put back $u={{e}^{x}}$ in the above step to have the integral;
\[\Rightarrow \sec \left( {{e}^{x}} \right)+c\]

Note: We note that we can prove $\int{\sec x\tan x}dx=\sec x+c$ by converting $\sec x\tan x$ into $\sin \cos x$ and then taking $\cos x=u$. We note that ${{e}^{x}}$ is the function whose integration and differentiation can return original function. We must be careful of the sign confusion which arises because of the similar formula $\int{\operatorname{cosec}x\cot x}dx=-\operatorname{cosec}x+c$.