Question

If $\int{\sqrt{1+\sec x}dx}=2\left( fog \right)\left( x \right)+C$, then
(a) $f\left( x \right)=\sec x-1$
(b) $f\left( x \right)=2{{\tan }^{-1}}x$
(c) $g\left( x \right)=\sqrt{\sec x-1}$
(d) None of these

Answer 1

Hint: Assume the integral in the L.H.S as $I$ and evaluate it. To do so assume $\left( 1+\sec x \right)={{k}^{2}}$ and differentiate both the sides to find the value of $dx$ in terms of $dk$. Convert all the functions present in terms of x into the terms of k using the identity ${{\sec }^{2}}x=1+{{\tan }^{2}}x$. Now, assume $k=\sqrt{2}\sec \theta $ differentiate both the sides to find the value of $d\theta $ in terms of $dk$. Further, assume $\tan \theta =m$ and change all the terms inside the integral in variable m. Finally, apply the formula $\int{\dfrac{dm}{{{m}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{m}{a} \right)$ to get the required integral. Substitute back all the assumed terms back into the variable x and get the composite function to find the correct option. Use the formulas: $\dfrac{d\left( \sec x \right)}{dx}=\sec x\tan x$ and $\dfrac{d\left( \tan \theta \right)}{d\theta }={{\sec }^{2}}\theta $.

Complete step by step solution:
Here we have been provided with the relation $\int{\sqrt{1+\sec x}dx}=2\left( fog \right)\left( x \right)+C$ and we are asked to check the functions $f\left( x \right)$ and $g\left( x \right)$ as per the options given. First we need to evaluate the integral in the L.H.S. Let us assume this integral as $I$ so we have,
$\Rightarrow I=\int{\sqrt{1+\sec x}dx}$
Assuming $\left( 1+\sec x \right)={{k}^{2}}$ and differentiating both the sides to find the value of $dx$ in terms of $dk$ we get,
\[\Rightarrow \dfrac{d\left( 1+\sec x \right)}{dx}=\dfrac{d\left( {{k}^{2}} \right)}{dx}\]
Using the formula $\dfrac{d\left( \sec x \right)}{dx}=\sec x\tan x$ in the L.H.S and the chain rule of differentiation in the R.H.S we get,
\[\begin{align}
  & \Rightarrow \sec x\tan x=\dfrac{d\left( {{k}^{2}} \right)}{dk}\times \dfrac{dk}{dx} \\
 & \Rightarrow \left( {{k}^{2}}-1 \right)\tan xdx=2kdk \\
 & \Rightarrow dx=\dfrac{2k}{\left( {{k}^{2}}-1 \right)\times \tan x}dk.........\left( i \right) \\
\end{align}\]
Now we need to find the value of $\tan x$ in terms of k to move ahead. We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ so we have,
$\begin{align}
  & \Rightarrow \tan x=\sqrt{{{\sec }^{2}}x-1} \\
 & \Rightarrow \tan x=\sqrt{{{\left( {{k}^{2}}-1 \right)}^{2}}-1} \\
\end{align}$
Using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ we get,
$\begin{align}
  & \Rightarrow \tan x=\sqrt{\left( {{k}^{2}}-2 \right)\left( {{k}^{2}} \right)} \\
 & \Rightarrow \tan x=k\sqrt{{{k}^{2}}-2} \\
\end{align}$
Substituting the above value in relation (i) we get,
 \[\begin{align}
  & \Rightarrow dx=\dfrac{2k}{\left( {{k}^{2}}-1 \right)\times k\sqrt{{{k}^{2}}-2}}dk \\
 & \Rightarrow dx=\dfrac{2}{\left( {{k}^{2}}-1 \right)\sqrt{{{k}^{2}}-2}}dk \\
\end{align}\]
Substituting the above relation in the integral $I$ we get,
\[\begin{align}
  & \Rightarrow I=\int{\dfrac{2\sqrt{{{k}^{2}}}}{\left( {{k}^{2}}-1 \right)\sqrt{{{k}^{2}}-2}}dk} \\
 & \Rightarrow I=2\int{\dfrac{k}{\left( {{k}^{2}}-1 \right)\sqrt{{{k}^{2}}-2}}dk} \\
\end{align}\]
Now, assuming $k=\sqrt{2}\sec \theta $ and differentiating both the sides using the formula $\dfrac{d\left( \sec \theta \right)}{d\theta }=\sec \theta \tan \theta $ we get,
\[\begin{align}
  & \Rightarrow \dfrac{dk}{d\theta }=\dfrac{d\left( \sqrt{2}\sec \theta \right)}{d\theta } \\
 & \Rightarrow dk=\left( \sqrt{2}\sec \theta \tan \theta \right)d\theta \\
\end{align}\]
Substituting this value in the integral we get,
\[\begin{align}
  & \Rightarrow I=2\int{\dfrac{\sqrt{2}\sec \theta \times \sqrt{2}\sec \theta \tan \theta }{\left( 2{{\sec }^{2}}\theta -1 \right)\sqrt{2{{\sec }^{2}}\theta -2}}d\theta } \\
 & \Rightarrow I=4\int{\dfrac{{{\sec }^{2}}\theta \tan \theta }{\left( 2{{\sec }^{2}}\theta -1 \right)\sqrt{2\left( {{\sec }^{2}}\theta -1 \right)}}d\theta } \\
 & \Rightarrow I=4\int{\dfrac{{{\sec }^{2}}\theta \tan \theta }{\left( 2{{\sec }^{2}}\theta -1 \right)\sqrt{2}\tan \theta }d\theta } \\
\end{align}\]
Cancelling the common terms and converting the secant function in the denominator into the tangent function we get,
\[\begin{align}
  & \Rightarrow I=2\sqrt{2}\int{\dfrac{{{\sec }^{2}}\theta }{\left( 2\left( 1+{{\tan }^{2}}\theta \right)-1 \right)}d\theta } \\
 & \Rightarrow I=2\sqrt{2}\int{\dfrac{{{\sec }^{2}}\theta }{2+2{{\tan }^{2}}\theta -1}d\theta } \\
 & \Rightarrow I=2\sqrt{2}\int{\dfrac{{{\sec }^{2}}\theta }{2{{\tan }^{2}}\theta +1}d\theta } \\
\end{align}\]
Taking 2 common from the denominator we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{2\sqrt{2}}{2}\int{\dfrac{{{\sec }^{2}}\theta }{{{\tan }^{2}}\theta +\dfrac{1}{2}}d\theta } \\
 & \Rightarrow I=\sqrt{2}\int{\dfrac{{{\sec }^{2}}\theta }{{{\tan }^{2}}\theta +{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}d\theta } \\
\end{align}\]
Substituting $\tan \theta =m$ and differentiating both the sides using the formula $\dfrac{d\left( \tan \theta \right)}{d\theta }={{\sec }^{2}}\theta $ we get,
$\begin{align}
  & \Rightarrow \dfrac{d\left( \tan \theta \right)}{d\theta }=\dfrac{dm}{d\theta } \\
 & \Rightarrow {{\sec }^{2}}\theta =\dfrac{dm}{d\theta } \\
 & \Rightarrow {{\sec }^{2}}\theta d\theta =dm \\
\end{align}$
Substituting the above relation in the integral we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{2\sqrt{2}}{2}\int{\dfrac{{{\sec }^{2}}\theta }{{{\tan }^{2}}\theta +\dfrac{1}{2}}d\theta } \\
 & \Rightarrow I=\sqrt{2}\int{\dfrac{dm}{{{m}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}} \\
\end{align}\]
Using the integral formula $\int{\dfrac{dm}{{{m}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{m}{a} \right)$ we get,
\[\begin{align}
  & \Rightarrow I=\sqrt{2}\times \dfrac{1}{\dfrac{1}{\sqrt{2}}}{{\tan }^{-1}}\left( \dfrac{m}{\dfrac{1}{\sqrt{2}}} \right) \\
 & \Rightarrow I=2{{\tan }^{-1}}\left( \sqrt{2}m \right)+C \\
\end{align}\]
Here C is the constant of integration. Substituting back the values of all the assumed variables $k,\theta $ and $m$ to get the function in terms of x and simplifying we get,
\[\Rightarrow I=2{{\tan }^{-1}}\left( \sqrt{\sec x-1} \right)+C\]
On comparing the above obtained integral with $2\left( fog \right)\left( x \right)+C$ which can be written as $2f\left( g\left( x \right) \right)+C$ we can conclude that the two functions are:
$\Rightarrow f\left( x \right)={{\tan }^{-1}}x$ and $g\left( x \right)=\sqrt{\sec x-1}$

So, the correct answer is “Option c”.

Note: Note that there are many ways to evaluate the integral and it may be possible that you may get the answer in terms of other trigonometric and inverse trigonometric functions. In the options we have the inverse tangent function as $f\left( x \right)$ and the secant function as $g\left( x \right)$ so even if we will get the value of integral in terms of other functions we will try to convert it terms of the functions given in the options and accordingly choose the correct one.