Answer
Verified
423k+ 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.
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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE