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Evaluate\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx\], where \[a > x\]

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Last updated date: 19th Jul 2024
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Answer
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Hint: We are given here to integrate the function \[\dfrac{1}{{({a^2} - {x^2})}}\], where \[a > x\]. To do this, first of all we will expand the denominator using the formula of \[({a^2} - {b^2})\]. Then we will multiply and divide the numerator and denominator by \[2a\]. Now we will write the numerator as \[a + x + a - x\]. Now we will solve further and get the value of this integration.
Formula used: We use the following formulas here,
\[\ln a - \ln b = \ln \dfrac{a}{b}\]
\[\int {\dfrac{1}{x}dx = \ln x + c} \]
\[({a^2} - {b^2}) = (a + b)(a - b)\]

Complete step-by-step solution:
We are given to integrate\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx\].
We know that \[({a^2} - {b^2}) = (a + b)(a - b)\]. Using this formula, we write the function \[\dfrac{1}{{({a^2} - {x^2})}}\] as,
\[\dfrac{1}{{({a^2} - {x^2})}} = \dfrac{1}{{(a + x)(a - x)}}\]
We multiply and divide the numerator and denominator by \[2a\].
\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{2a}}{{2a}} \cdot \dfrac{1}{{(a + x)(a - x)}}\]
We now add and subtract \[x\] on the numerator as,
\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{2a}}{{2a}} \cdot \dfrac{{x - x}}{{(a + x)(a - x)}}\]
We will now rearrange this function,
\[ \Rightarrow \dfrac{1}{{({a^2} - {x^2})}} = \dfrac{{(a + x) + (a - x)}}{{2a(a + x)(a - x)}}\]
We will use this step in the integration. We replace \[\dfrac{1}{{({a^2} - {x^2})}}\] with RHS of above step as,
\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\dfrac{{(a + x) + (a - x)}}{{(a + x)(a - x)}}dx} \]
On simplifying further,
\[
   \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\left( {\dfrac{{a + x}}{{(a + x)(a - x)}}} \right)dx + \dfrac{1}{{2a}}\int {\left( {\dfrac{{a - x}}{{(a + x)(a - x)}}} \right)dx} } \\
   \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\int {\left( {\dfrac{1}{{(a - x)}}} \right)dx + \dfrac{1}{{2a}}\int {\left( {\dfrac{1}{{(a + x)}}} \right)dx} } \\
\]
We know that \[\int {\dfrac{1}{x}dx = \ln x + c} \], using this above we get,
\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left( { - \ln (a - x)} \right) + \dfrac{1}{{2a}}\left( {\ln \left( {a + x} \right)} \right)\]
Since, \[a > x\] we can safely put the value \[a - x\] in log function as \[a - x > 0\].
\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left[ {\ln \left( {a + x} \right) - \ln (a - x)} \right]\]
As we know that \[\ln a - \ln b = \ln \dfrac{a}{b}\], we move ahead as,
\[ \Rightarrow \int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \dfrac{1}{{2a}}\left[ {\ln \left( {\dfrac{{a + x}}{{a - x}}} \right)} \right]\]
Hence we have got the value of the given integration as \[\dfrac{1}{{2a}}\left[ {\ln \left( {\dfrac{{a + x}}{{a - x}}} \right)} \right]\].

Note: This is to note that we could have also done this integration by the method of partial fraction. But that method is a bit longer. Moreover, if you are not able to find the suitable modifications on the function you can always use a partial fraction method in such types of questions. In partial fraction, we will write the integration as,
\[\int {\dfrac{1}{{({a^2} - {x^2})}}} dx = \int {\left[ {\dfrac{A}{{a + x}} + \dfrac{B}{{a - x}}} \right]dx} \]
We then find the value of \[A\] and \[B\] and then solve ahead.