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How can you proof $\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c$ using $x = a\sin \theta$?

Last updated date: 13th Jun 2024
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Hint: Substitute the given value for $x$ in the left hand side of the integration and replace $dx$ accordingly. Simplify the expression in integration using formula $1 - {\sin ^2}\theta = {\cos ^2}\theta$ and then integrate it. Determine the other trigonometric ratios from $x = a\sin \theta$ to get the integration result in terms of $x$. Simplify the final result and bring it in the form of right hand side.

According to the question, we have to prove the integration using the given substitution.
The integration to prove is:
$\Rightarrow \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c{\text{ }}.....{\text{(1)}}$
Let the left hand side integral is denoted as $I$, then we have:
$\Rightarrow I = \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} {\text{ }}.....{\text{(2)}}$
Now as it is given that we have to use substitution to prove it. So we have:
$\Rightarrow x = a\sin \theta$
Differentiating it both sides, we’ll get:
$\Rightarrow dx = a\cos \theta d\theta$
Putting these values in integration equation (2), we’ll get:
$\Rightarrow I = \int {\dfrac{{a\cos \theta d\theta }}{{{a^2} - {{\left( {a\sin \theta } \right)}^2}}}}$
Simplifying it further, we’ll het:
$\Rightarrow I = \int {\dfrac{{a\cos \theta d\theta }}{{{a^2} - {a^2}{{\sin }^2}\theta }}} \\ \Rightarrow I = \int {\dfrac{{a\cos \theta d\theta }}{{{a^2}\left( {1 - {{\sin }^2}\theta } \right)}}} \\$
We know the trigonometric formula $1 - {\sin ^2}\theta = {\cos ^2}\theta$. Using this, we’ll get:
$\Rightarrow I = \int {\dfrac{{\cos \theta d\theta }}{{a{{\cos }^2}\theta }}} \\ \Rightarrow I = \dfrac{1}{a}\int {\dfrac{{d\theta }}{{\cos \theta }}} \\ \Rightarrow I = \dfrac{1}{a}\int {\sec \theta d\theta } \\$
We know the integration formula $\int {\sec x = } \log \left| {\sec x + \tan x} \right| + c$. Using this formula, we’ll get:
$\Rightarrow I = \dfrac{1}{a}\log \left| {\sec \theta + \tan \theta } \right| + c{\text{ }}.....{\text{(3)}}$
We have used $x = a\sin \theta$. From this we have:
$\Rightarrow \sin \theta = \dfrac{x}{a}$
Using the value of $\sin \theta$, we can determine other trigonometric ratios. So we have:
$\Rightarrow \sec \theta = \dfrac{a}{{\sqrt {{a^2} - {x^2}} }}$ and $\tan \theta = \dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$. Putting these values in equation (3), we’ll get:
$\Rightarrow I = \dfrac{1}{a}\log \left| {\dfrac{a}{{\sqrt {{a^2} - {x^2}} }} + \dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right| + c$
Simplifying this further, we’ll get:
$\Rightarrow I = \dfrac{1}{a}\log \left| {\dfrac{{\left( {a + x} \right)}}{{\sqrt {{a^2} - {x^2}} }}} \right| + c$
Using the algebraic formula $\left( {{x^2} - {a^2}} \right) = \left( {x - a} \right)\left( {x + a} \right)$, we’ll get:
$\Rightarrow I = \dfrac{1}{a}\log \left| {\dfrac{{\left( {a + x} \right)}}{{\sqrt {\left( {a + x} \right)\left( {a - x} \right)} }}} \right| + c \\ \Rightarrow I = \dfrac{1}{a}\log \left| {\dfrac{{\left( {a + x} \right)}}{{\sqrt {\left( {a + x} \right)} \sqrt {\left( {a - x} \right)} }}} \right| + c \\ \Rightarrow I = \dfrac{1}{a}\log \left| {\dfrac{{\sqrt {\left( {a + x} \right)} }}{{\sqrt {\left( {a - x} \right)} }}} \right| + c \\ \Rightarrow I = \dfrac{1}{a}\log {\left| {\dfrac{{a + x}}{{a - x}}} \right|^{\dfrac{1}{2}}} + c \\$
Applying the logarithmic formula $\log {a^b} = b\log a$, we’ll get:
$\Rightarrow I = \dfrac{1}{a} \times \dfrac{1}{2}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c \\ \Rightarrow I = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c \\$
Putting the value of $I$ from equation (2), we’ll get:
$\Rightarrow \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c$
This is the required proof of the integration.

Note: The integration can also be by partial fraction method as shown:
$\Rightarrow \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \int {\dfrac{{dx}}{{\left( {a - x} \right)\left( {a + x} \right)}}}$
Now we can apply partial fraction, the expression in the integration can be written as:
$\Rightarrow \dfrac{1}{{\left( {a - x} \right)\left( {a + x} \right)}} = \dfrac{1}{{2a}}\left( {\dfrac{1}{{a + x}} + \dfrac{1}{{a - x}}} \right)$
Using this partial fraction in the above integration, we’ll get:
$\Rightarrow \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \int {\dfrac{1}{{2a}}\left( {\dfrac{1}{{a + x}} + \dfrac{1}{{a - x}}} \right)dx} \\ \Rightarrow \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\left( {\int {\dfrac{{dx}}{{a + x}} + \int {\dfrac{{dx}}{{a - x}}} } } \right) \\$
Now we can easily integrate this and we will get the same result.