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Question: Evaluate the indefinite integral of \(\int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx}\) where function...

Evaluate the indefinite integral of ∫extan⁡exsec⁡exdx\int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx} where functions are well defined. $$$$

Explanation

Solution

We recall the definition of indefinite integration and the substitution by integration method to solve the indefinite integration. We in order to substitute ex=u{{e}^{x}}=u in the integrand of the function find dudu and then substituteex=u{{e}^{x}}=u. We use the standard integration ∫sec⁡xtan⁡xdx=sec⁡x+c\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 ff is a differentiable function FF whose derivative is equal to the original function ff which meansF′=f{{F}^{'}}=f. The process of finding integral is called integration and the original function is called integrand. We write integration in variable xx as
∫f(x)dx=F(x)+c\int{f\left( x \right)}dx=F\left( x \right)+c
Here cc is an arbitrary constant of integration. We also integral remains the same even if we change the variable. It means for any variable uu we have;
∫f(u)du=∫f(u)du\int{f\left( u \right)}du=\int{f\left( u \right)}du
If we have composite function f(g(x))f\left( g\left( x \right) \right)and the differential of the function inside the bracket g′(x){{g}^{'}}\left( x \right)we can substitute the g(x)g\left( x \right) as by variable uu which meansg(x)=ug\left( x \right)=uwe can integrate as
∫f(g(x))g′(x)dx=∫f(u)du\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∫extan⁡exsec⁡exdx\int{{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}dx}. We see that here the integrand is extan⁡exsec⁡ex{{e}^{x}}\tan {{e}^{x}}\sec {{e}^{x}}. We see that the function ex{{e}^{x}} is repeated 3 times and as well as we know that ddxex=ex\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}. So we solve the problem by the u-substitution method. We in order to substitute ex=u{{e}^{x}}=u in the integrand of the function find dudu by differentiating with respect to uu. We have;

& \dfrac{d}{dt}{{e}^{x}}=\dfrac{du}{dx} \\\ & \Rightarrow {{e}^{x}}=\dfrac{du}{dx} \\\ & \Rightarrow du={{e}^{x}}dx \\\ \end{aligned}$$ Now we put $u={{e}^{x}},du={{e}^{x}}dx$ in the integrand to have; $$\begin{aligned} & \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{aligned}$$ We use the standard integration $\int{\sec x\tan x}dx=\sec x+c$ in the above step to have; $$\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 $\int{\operatorname{cosec}x\cot x}dx=-\operatorname{cosec}x+c$.