Question

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

Answer 1

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.

Complete step by step answer:
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.