# Evaluate the following integral:

\[\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}\]

Answer

Verified

381.3k+ views

**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,

\[{{e}^{x}}dx=du\]

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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

Which one of the following places is unlikely to be class 8 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference Between Plant Cell and Animal Cell

Find the HCF and LCM of 6 72 and 120 using the prime class 6 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers