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Last updated date: 04th Dec 2023
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MVSAT Dec 2023

Evaluate the following integral:
\[\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}\]

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Hint: As we have no direct formula for the given integral. Therefore, substitute \[{{e}^{x}}=u\] and express integral in terms of u. Then again substitute \[u=\sec \theta \] and then finally solve the integral to get the required value.

Complete step-by-step answer:
Here, we have to solve the integral \[\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}\].
Let us consider the integral given in the question,
\[I=\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}\]
Let us take \[{{e}^{x}}=u\]
We know that \[\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\]
Hence, by differentiating both sides, we get,
Or, \[u\text{ }dx=du\]
\[\Rightarrow dx=\dfrac{du}{u}\]
By substituting the value of \[{{e}^{x}}=u\] and \[dx=\dfrac{du}{u}\] in the given integral, we get,
\[I=\int{\sqrt{{{\left( u \right)}^{2}}-1}\dfrac{du}{u}}\]
Or, \[I=\int{\dfrac{\sqrt{{{u}^{2}}-1}}{u}du.....\left( i \right)}\]
Now, let us take \[u=\sec \theta \]
We know that \[\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x\]
Therefore by differentiating both sides, we get
\[du=\sec \theta \tan \theta d\theta \]
Now, by substituting \[u=\sec \theta \] and \[du=\sec \theta \tan \theta d\theta \] in equation (i), we get,
\[I=\int{\dfrac{\sqrt{{{\sec }^{2}}\theta -1}}{\sec \theta }.\sec \theta .\tan \theta \text{ }d\theta }\]
Now, by canceling like terms and substituting \[{{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \] in the above expression, we get
\[I=\int{\sqrt{{{\tan }^{2}}\theta }.\tan \theta d\theta }\]
Or, \[I=\int{{{\tan }^{2}}\theta \text{ }d\theta }\]
Again, we know that \[{{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1\]. Therefore we get,
\[I=\int{\left( {{\sec }^{2}}\theta -1 \right)d\theta }\]
We know that \[\int{{{\sec }^{2}}x\text{ }dx=\tan x}\] and \[\int{kdx=kx}\]
By using these, we get,
\[I=\tan \theta -\theta +C....\left( ii \right)\]
We know that \[u=\sec \theta \], so we get,
\[\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}\]
\[\tan \theta =\sqrt{{{u}^{2}}-1}\]
Also, \[\theta ={{\sec }^{-1}}\left( u \right)\]
By substituting the value of \[\tan \theta \] and \[\theta \] in equation (ii), we get,
\[I=\sqrt{{{u}^{2}}-1}-{{\sec }^{-1}}\left( u \right)+C\]
Now by substituting \[u={{e}^{x}}\] in the above equation, we get,
\[I=\sqrt{{{\left( {{e}^{x}} \right)}^{2}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C\]
Or, \[I=\sqrt{{{e}^{2x}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C\]
Therefore, we get \[\int{\sqrt{{{e}^{2x}}-1}dx=\sqrt{{{e}^{2x}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C}\]

Note: Here, students can cross-check their answers by differentiating the answer and checking if it is giving the original integral or not. Also, students can remember these substitutions in these cases to easily solve the questions of this type. Also, students should always convert the answer of the given integral in its original variable like here we have converted \[\theta \] back to x at the end.