How do you integrate $ \int {\dfrac{1}{{{{\left( {{{(ax)}^2} - {b^2}} \right)}^{\dfrac{3}{2}}}}}} $ by trigonometric substitution?

Hint: Start by substituting $ x = \dfrac{b}{a}\sec \theta $ . Substitute the values in place of the terms to make the equation easier to solve. Then we will differentiate the term. Hence, the differentiated term will be $ dx = \dfrac{b}{a}\sec \theta \tan \theta d\theta $ . Now we will substitute these terms in the original expression and integrate.

Complete step-by-step answer:
First we will start off by substituting $ x = \dfrac{b}{a}\sec \theta $ . Now we differentiate this term to form a proper equation and for substituting the terms.
 $
  \,\,x = \dfrac{b}{a}\sec \theta \\
  dx = \dfrac{b}{a}\sec \theta \tan \theta d\theta \\
  $
Now, we substitute the terms in our original expression.
\[
   = \int {\dfrac{1}{{{{\left( {{{(ax)}^2} - {b^2}} \right)}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{\dfrac{b}{a}\sec \theta \tan \theta d\theta }}{{{{\left( {({a^2})\left( {\dfrac{{{b^2}}}{{{a^2}}}{{\sec }^2}\theta } \right) - {b^2}} \right)}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{\dfrac{b}{a}\sec \theta \tan \theta d\theta }}{{{{\left( {\left( {{b^2}{{\sec }^2}\theta } \right) - {b^2}} \right)}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{b\sec \theta \tan \theta d\theta }}{{{{\left( {a\left( {{b^2}{{\sec }^2}\theta } \right) - {b^2}} \right)}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{b\sec \theta \tan \theta d\theta }}{{a{{\left( {{b^2}} \right)}^{\dfrac{3}{2}}}{{({{\sec }^2}\theta - 1)}^{\dfrac{3}{2}}}}}} \\
 \]
Now, we know that $ 1 + {\tan ^2}\theta = {\sec ^2}\theta $ .
Hence, we can also write it as $ {\tan ^2}\theta = {\sec ^2}\theta - 1 $ .
Therefore, the expression will become,
\[
   = \int {\dfrac{{b\sec \theta \tan \theta d\theta }}{{a{{\left( {{b^2}} \right)}^{\dfrac{3}{2}}}{{({{\sec }^2}\theta - 1)}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{b\sec \theta \tan \theta d\theta }}{{\left( {a{b^3}} \right){{({{\tan }^2}\theta )}^{\dfrac{3}{2}}}}}} \\
   = \int {\dfrac{{\sec \theta \tan \theta d\theta }}{{\left( {a{b^2}} \right){{({{\tan }^2}\theta )}^{\dfrac{3}{2}}}}}} \\
   = \dfrac{1}{{a{b^2}}}\int {\dfrac{{\sec \theta \tan \theta d\theta }}{{({{\tan }^3}\theta )}}} \\
   = \dfrac{1}{{a{b^2}}}\int {\dfrac{{\sec \theta d\theta }}{{({{\tan }^2}\theta )}}} \\
 \]
Now here we will substitute,
  $
  \sec \theta = \dfrac{1}{{\cos \theta }} \\
  \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \\
  $
So, the equation becomes,
\[
   = \dfrac{1}{{a{b^2}}}\int {\dfrac{{\sec \theta d\theta }}{{({{\tan }^2}\theta )}}} \\
   = \dfrac{1}{{a{b^2}}}\int {\dfrac{1}{{\cos \theta }}\left( {\dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \\
   = \dfrac{1}{{a{b^2}}}\int {\left( {\dfrac{{\cos \theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \\
 \]
Now, here consider $ u = \sin \theta $ and $ du = \cos \theta d\theta $ .
Hence, the expression will become,
 $ = \dfrac{1}{{a{b^2}}}\int {\dfrac{{du}}{{{u^2}}}} $
Now we integrate $ {u^{ - 2}} $ by using the power rule for integration which will gives us,
 $ = \dfrac{1}{{a{b^2}}}\left( { - \dfrac{1}{u}} \right) $
Now substitute the value of $ u $ .
Hence, the equation becomes,
 $ = - \dfrac{1}{{a{b^2}}}\cos ec\left( {\cos \left( {\dfrac{{ax}}{b}} \right)} \right) $
So, the correct answer is “ $ - \dfrac{1}{{a{b^2}}}\cos ec\left( {\cos \left( {\dfrac{{ax}}{b}} \right)} \right) $ ”.

Note: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
While substituting the terms make sure you are taking into account the degrees and signs of the terms as well. While applying the power rule make sure you have considered the power with their respective signs. Remember the identity $ 1 + {\tan ^2}\theta = {\sec ^2}\theta $ .