Question

Find the integral of $ \dfrac{1}{{\sqrt {{x^2} - {a^2}} }} $ with respect to $ x $ and hence evaluate $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} $

Answer 1

Hint: Here first of all we will derive the formula for $ \dfrac{1}{{\sqrt {{x^2} - {a^2}} }} $ and then will compare the given expression $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} $ to evaluate with it and then find the resultant required value for it using the formulas.

Complete step by step solution:
Take the expression: $ \dfrac{1}{{\sqrt {{x^2} - {a^2}} }} $ …. (A)
Now, take the part of the above expression
Let us assume that $ x = a\sec \theta $
Differentiate the above expression with respect to
 $ \dfrac{{dx}}{{d\theta }} = a\sec \theta \tan \theta $
The above expression can be written as –
 $ dx = a\sec \theta \tan \theta d\theta $ ….. (B)
Now, $ \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}{{\sec }^2}\theta - {a^2}} $
Simplify the above expression using the identity
\[
  \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}({{\sec }^2}\theta - 1)} \\
  \sqrt {{x^2} - {a^2}} = \sqrt {{a^2}({{\tan }^2}\theta )} \\
  \sqrt {{x^2} - {a^2}} = a\tan \theta \;{\text{ }}.....{\text{ (C)}} \;
 \]
Therefore, the equation (A) can be re-written by using the equations (B) and (c)
 $ I = \int {\dfrac{{a\sec \theta \tan \theta d\theta }}{{a\tan \theta }}} $
Common factors from the numerator and the denominator cancel each other. Therefore, remove from the numerator and the denominator of the above equation.
 $ I = \int {\sec \theta d\theta } $
By using the formula of Integration-
 $ I = \int {\sec \theta d\theta } = \ln \left| {\sec \theta + \tan \theta } \right| + C $
Replace the values in the above expression
 $ I = \int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx} = \ln \left( {\dfrac{x}{a} + \sqrt {{{\left( {\dfrac{x}{a}} \right)}^2} - 1} } \right) + C $ …. (D)
Now, Integration for the required expression $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} $ by comparing with the above expression
 $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} = \int {\dfrac{1}{{\sqrt {{x^2} - {5^2}} }}dx} $
Placing the Integration in the above expression –
 $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} = \ln \left( {\dfrac{x}{5} + \sqrt {\dfrac{{{x^2}}}{{25}} - 1} } \right) + C $
This is the required solution.
So, the correct answer is “ $ \int {\dfrac{1}{{\sqrt {{x^2} - 25} }}dx} = \ln \left( {\dfrac{x}{5} + \sqrt {\dfrac{{{x^2}}}{{25}} - 1} } \right) + C $ ”.

Note: Anti-derivative is another name of the inverse derivative, the primitive function and the primitive integral or the indefinite integral of a function f is the differentiable function F whose derivative is equal to the original function f. Know the difference between the differentiation and the integration and apply formula and the properties accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other.