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
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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.
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.
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