Question

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

Answer 1

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}}} \end{aligned}$$ Now since $$\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$$, therefore the above integral becomes $$\begin{aligned} & {{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{aligned}$$ We also have the integral $${{I}_{1}}=\int {{e}^{x}}\tan x\sec xdx$$, hence using the value of $${{I}_{1}}$$ in the above integral, we have $$\begin{aligned} & {{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{aligned}$$ Now substituting the value of integral $${{I}_{1}}$$ and $${{I}_{2}}$$ in integral $$I$$, we get $$\begin{aligned} & 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{aligned}$$ We are also given that $$\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.