Question

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 $f\left( x \right)$ is
A.$\sec x+x\tan x+\dfrac{1}{2}$
B.$x\sec x+x\tan x+\dfrac{1}{2}$
C.$x\sec x+{{x}^{2}}\tan x+\dfrac{1}{2}$
D.$\sec x+\tan x-x+\dfrac{1}{2}$

Answer 1

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.