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# If $\int{{{e}^{\sec x}}\left( \sec x\tan xf\left( x \right)+\sec x\tan x+{{\tan }^{2}}x \right).dx={{e}^{\sec x}}f\left( x \right)+c}$. Then $f\left( x \right)$ isA.$\sec x+x\tan x+\dfrac{1}{2}$ B.$x\sec x+x\tan x+\dfrac{1}{2}$ C.$x\sec x+{{x}^{2}}\tan x+\dfrac{1}{2}$ D.$\sec x+\tan x-x+\dfrac{1}{2}$

Last updated date: 20th Jun 2024
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Hint: We are given a function which consists of multiple trigonometric functions. First we will differentiate the given equation which would help us to simplify our equation and get rid of ${{e}^{\sec x}}$. Then, we will find the anti-derivative or integral of the equation thus obtained to transform ${f}'\left( x \right)$ to $f\left( x \right)$.

Complete step by step solution:
We are given the equation $\int{{{e}^{\sec x}}\left( \sec x\tan xf\left( x \right)+\sec x\tan x+{{\tan }^{2}}x \right).dx={{e}^{\sec x}}f\left( x \right)+c}$.
Differentiating both sides, we get
$\Rightarrow \dfrac{d}{dx}\left\{ \int{{{e}^{\sec x}}\left( \sec x\tan xf\left( x \right)+\sec x\tan x+{{\tan }^{2}}x \right).dx} \right\}=\dfrac{d}{dx}\left( {{e}^{\sec x}}f\left( x \right)+c \right)$
\begin{align} & \Rightarrow {{e}^{\sec x}}\left( \sec x\tan xf\left( x \right)+\sec x\tan x+{{\tan }^{2}}x \right)={{e}^{\sec x}}{f}'\left( x \right)+{{e}^{\sec x}}\sec x\tan xf\left( x \right) \\ & \Rightarrow {{e}^{\sec x}}\sec x\tan xf\left( x \right)+{{e}^{\sec x}}\sec x\tan x+{{e}^{\sec x}}{{\tan }^{2}}x={{e}^{\sec x}}{f}'\left( x \right)+{{e}^{\sec x}}\sec x\tan xf\left( x \right) \\ \end{align}
Cancelling ${{e}^{\sec x}}\sec x\tan xf\left( x \right)$ from both sides, we get
\begin{align} & \Rightarrow {{e}^{\sec x}}\sec x\tan x+{{e}^{\sec x}}{{\tan }^{2}}x={{e}^{\sec x}}{f}'\left( x \right) \\ & \Rightarrow {{e}^{\sec x}}\left( \sec x\tan x+{{\tan }^{2}}x \right)={{e}^{\sec x}}{f}'\left( x \right) \\ \end{align}
Now, dividing both sides by ${{e}^{\sec x}}$, we get
$\Rightarrow \sec x\tan x+{{\tan }^{2}}x={f}'\left( x \right)$
Here, in order to find $f\left( x \right)$, we shall calculate the antiderivative or the integral of ${f}'\left( x \right)$.
\begin{align} & \Rightarrow \int{{f}'\left( x \right).dx=\int{\left( \sec x\tan x+{{\tan }^{2}}x \right).dx}} \\ & \Rightarrow \int{{f}'\left( x \right).dx=\int{\sec x\tan x.dx}}+\int{{{\tan }^{2}}x.dx} \\ \end{align}
We shall calculate these two integrals individually and then combine them to find our final result.
Since, $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$ ,
\begin{align} & \Rightarrow \int{\sec x\tan x.dx}=\int{\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x}.dx} \\ & \Rightarrow \int{\sec x\tan x.dx}=\int{\dfrac{\sin x}{{{\left( \cos x \right)}^{2}}}.dx} \\ \end{align}
Performing simple substitution, we see that if $t=\cos x$.
Then, $dt=-\sin x.dx$
\begin{align} & \Rightarrow \int{\sec x\tan x.dx}=\int{\dfrac{1}{{{\left( t \right)}^{2}}}.-dt} \\ & \Rightarrow \int{\sec x\tan x.dx}=-\int{{{t}^{-2}}dt} \\ \end{align}
Using the property of integration, $\int{{{x}^{n}}.dx=}\dfrac{{{x}^{n+1}}}{n+1}+C$, we get
\begin{align} & \Rightarrow \int{\sec x\tan x.dx}=-\dfrac{{{t}^{-2+1}}}{-2+1}+C \\ & \Rightarrow \int{\sec x\tan x.dx}=\dfrac{1}{t}+C \\ \end{align}
Substituting the value of $t=\cos x$, we get
$\Rightarrow \int{\sec x\tan x.dx}=\dfrac{1}{\cos x}+C$
$\Rightarrow \int{\sec x\tan x.dx}=\sec x+C$ …………………. (1)
Also, we know that ${{\tan }^{2}}x={{\sec }^{2}}x-1$, substituting this value, we get
\begin{align} & \Rightarrow \int{{{\tan }^{2}}x.dx}=\int{\left( {{\sec }^{2}}x-1 \right)}.dx \\ & \Rightarrow \int{{{\tan }^{2}}x.dx}=\int{{{\sec }^{2}}x.dx}-\int{1.dx} \\ \end{align}
Using the property of integration, $\int{{{\sec }^{2}}.dx=}\tan x+C$ as well as $\int{1.dx=x+C}$ , we get$\Rightarrow \int{{{\tan }^{2}}x.dx}=\tan x+C-x+C$
$\Rightarrow \int{{{\tan }^{2}}x.dx}=\tan x-x+C$ ………………….. (2)
Combining (1) and (2), we get
\begin{align} & \Rightarrow \int{{f}'\left( x \right).dx=\left( \sec x+C \right)}+\left( \tan x-x+C \right) \\ & \Rightarrow f\left( x \right)+C=\sec x+\tan x-x+C \\ \end{align}
$\Rightarrow f\left( x \right)=\sec x+\tan x-x+C$
We shall now compare the calculated value of f(x) with the options given in the problem and we find that it best matches option (D).

Therefore, the correct option is (D) $\sec x+\tan x-x+\dfrac{1}{2}$.

Note:
While performing indefinite integration, we must take special care about adding the constant of integration, C to our answer of integral. If we substitute the values of any point lying on a particular curve, then we can calculate the exact value of this constant of integration and hence, the general equation transforms into the equation of that particular curve only.