Question

Integrate the given expression, \[\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}\] .

Answer 1

Hint: Assume \[t={{e}^{x}}\] . Use the equation \[dt={{e}^{x}}dx\] and then simplify the expression in terms of t as \[\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}\]. Consider \[t=\sec \theta \] . Use the equation \[dt=\sec \theta .\tan \theta d\theta \] , and then simplify the expression \[\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }\] . We know that, \[\int{secxdx}=ln\left| secx\left. +tanx \right| \right.\] . Use this formula and solve the expression further. Then, express \[\theta \] in terms of t. And then, express t in terms of x as we assumed \[t={{e}^{x}}\] .

Complete step-by-step solution -
Let us assume, \[t={{e}^{x}}\]
\[{{t}^{2}}={{e}^{2x}}\] …………….(1)
Differentiating equation (1) with respect to x, we get
\[\dfrac{dt}{dx}={{e}^{x}}\]
\[\Rightarrow dt={{e}^{x}}dx\] ………………..(2)
Now, using equation (1) and equation (2), we can transform the expression \[\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}\] .
Transforming the expression, we get \[\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}\] ……….(3)
Now, let us assume,
\[t=\sec \theta \] …………..(4)
Differentiating equation (4) with respect to \[\theta \] , we get
\[\dfrac{dt}{d\theta }=\sec \theta .\tan \theta \]
\[\Rightarrow dt=\sec \theta .\tan \theta d\theta \] ………………..(5)
Now, using equation (5), we can transform equation (3)
\[\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}\]
\[=\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }\] ……………….(6)
We know the identity, \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]
\[\Rightarrow {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \] ………………………(7)
Now, using equation (7), we can transform equation (6)
\[=\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }\]
\[\begin{align}
  & =\int{\dfrac{1}{\sqrt{{{\tan }^{2}}\theta }}\sec \theta .\tan \theta d\theta } \\
 & =\int{\dfrac{1}{\tan \theta }}.\sec \theta \tan \theta d\theta \\
\end{align}\]
\[=\int{\sec \theta d\theta }\] …………………(8)
We know that, \[\int{secxdx}=ln\left| secx\left. +tanx \right| \right.\] ………….(9)
Replacing x by \[\theta \] in the equation (9), we get
\[\int{secxdx}=ln\left| secx\left. +tanx \right| \right.\]
\[\Rightarrow \int{\sec \theta d\theta }=\ln \left| \sec \theta \left. +\tan \theta \right| \right.\] …………….(10)
From equation (4), we have
\[t=\sec \theta \]
We can find the value of \[\tan \theta \] using equation (7).
\[\begin{align}
  & {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\
 & \Rightarrow {{t}^{2}}-1={{\tan }^{2}}\theta \\
 & \Rightarrow \tan \theta =\sqrt{{{t}^{2}}-1} \\
\end{align}\]
Initially, we have assumed \[t={{e}^{x}}\]. Now, putting the value of t, we get
\[t=\sec \theta ={{e}^{x}}\]
\[\begin{align}
  & \tan \theta =\sqrt{{{t}^{2}}-1} \\
 & \Rightarrow \tan \theta =\sqrt{{{e}^{2x}}-1} \\
\end{align}\]
Now, putting the values of \[\sec \theta \] and \[\tan \theta \] in equation (10), we get
\[\Rightarrow \int{\sec \theta d\theta }=\ln \left| \sec \theta \left. +\tan \theta \right| \right.\]
\[=\ln \left| {{e}^{x}}\left. +\sqrt{{{e}^{2x}}-1} \right| \right.\]
Therefore, \[\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}=\ln \left| {{e}^{x}}\left. +\sqrt{{{e}^{2x}}-1} \right| \right.\] .

Note: In this question, one can think to assume \[{{e}^{2x}}\] as t in the expression \[\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}\] . But, if we do so then our equation will look like \[\int{\sqrt{t}.\dfrac{1}{\sqrt{{{t}^{2}}-1}}dx}\] . Here, we are unable to replace \[dx\] as \[dt\] . We can see that this expression has become complex to be solved further. So, this approach is not suitable for the integration of this expression.