
How do you evaluate the integral of \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}\] ?
Answer
556.8k+ views
Hint: In order to solve the above question, we have to apply trigonometric substitutions, first we have to make a few substitutions so that the given integral is simplified After that we will integrate term by term using simple integration formulas.
Complete step by step answer:
The above question belongs to the concept of integration by trigonometric substitution. Here we have to use basic trigonometric substitutions in order to integrate the given function. We have to evaluate \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}\].
We will first make a few substitutions.
Here we will use hyperbolic functions to make the substitution.
We know that \[{{\cosh }^{2}}z-{{\sinh }^{2}}z=1\]
Our first step is to let \[x=a\sinh y,dx=a\cosh ydy\]
In order to convert the derivative in our integral expression we have to manipulate the integral in terms of the substitution.
Now replacing variables with the substitution in the given integral and transforming the integral in terms of the substitution.
\[\begin{align}
& \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=\int{\dfrac{1}{\sqrt{{{a}^{2}}+{{a}^{2}}{{\sinh }^{2}}y}}a\cosh ydy} \\
& \Rightarrow \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=\int{\dfrac{1}{a\cosh y}a\cosh ydy} \\
& \Rightarrow \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=y+C \\
\end{align}\]
After applying the integration rule, we get.
\[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=y+C\]
Now reverse the substitution which we used
\[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}={{\sinh }^{-1}}\dfrac{x}{a}+C\]
As we know that \[{{\sinh }^{-1}}z=\ln \left( z+\sqrt{{{z}^{2}}+1} \right)\]
Therefore, the integration of the given integral \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}\] is \[\ln \left( \dfrac{x}{a}+\sqrt{{{\left( \dfrac{x}{a} \right)}^{2}}+1} \right)\] .
Note:
While solving the above question be careful with the integration part. Do remember the substitution method used here for future use. Try to solve the question step by step. To solve these types of questions, we should have the knowledge of trigonometric identities. Do not forget to reverse the substitution. After integration add the constant of integration in the result.
Complete step by step answer:
The above question belongs to the concept of integration by trigonometric substitution. Here we have to use basic trigonometric substitutions in order to integrate the given function. We have to evaluate \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}\].
We will first make a few substitutions.
Here we will use hyperbolic functions to make the substitution.
We know that \[{{\cosh }^{2}}z-{{\sinh }^{2}}z=1\]
Our first step is to let \[x=a\sinh y,dx=a\cosh ydy\]
In order to convert the derivative in our integral expression we have to manipulate the integral in terms of the substitution.
Now replacing variables with the substitution in the given integral and transforming the integral in terms of the substitution.
\[\begin{align}
& \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=\int{\dfrac{1}{\sqrt{{{a}^{2}}+{{a}^{2}}{{\sinh }^{2}}y}}a\cosh ydy} \\
& \Rightarrow \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=\int{\dfrac{1}{a\cosh y}a\cosh ydy} \\
& \Rightarrow \int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=y+C \\
\end{align}\]
After applying the integration rule, we get.
\[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}=y+C\]
Now reverse the substitution which we used
\[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}={{\sinh }^{-1}}\dfrac{x}{a}+C\]
As we know that \[{{\sinh }^{-1}}z=\ln \left( z+\sqrt{{{z}^{2}}+1} \right)\]
Therefore, the integration of the given integral \[\int{\dfrac{dx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}\] is \[\ln \left( \dfrac{x}{a}+\sqrt{{{\left( \dfrac{x}{a} \right)}^{2}}+1} \right)\] .
Note:
While solving the above question be careful with the integration part. Do remember the substitution method used here for future use. Try to solve the question step by step. To solve these types of questions, we should have the knowledge of trigonometric identities. Do not forget to reverse the substitution. After integration add the constant of integration in the result.
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