Question

How do you integrate $ \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx $ using trigonometric substitution?

Answer 1

Hint: To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value.

Complete step-by-step answer:
Let, \[I = \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx\] .
Now, we can substitute:
 $ x = 2.\sec y $
Differentiate the above equation,
 $
   \Rightarrow \dfrac{{dx}}{{dy}} = 2\sec y.\tan y \\
   \Rightarrow dx = 2\sec y.\tan y.dy \;
  $
Put the value of $ x $ in $ I $ :
 $
  \therefore I = \int \dfrac{{4{{\sec }^2}y}}{{\sqrt {4{{\sec }^2}y - 4} }}.2\sec y.\tan ydy \\
   = \int \dfrac{{4{{\sec }^2}y}}{{2\tan y}}.2\sec y.\tan ydy \\
   = 4\int {\sec ^3}ydy \\
   = 4\int \sec y.{\sec ^2}ydy \\
   = 4\int \sqrt {{{\tan }^2}y + 1} .{\sec ^2}ydy \;
  $
Now, we substitute: $ \tan y = t $ ;
Differentiate the above substitution:
 $ \Rightarrow {\sec ^2}ydy = dt $
 $
  \therefore I = 4\int \sqrt {{t^2} + 1} dt \\
   = 4\{ \dfrac{t}{2}\sqrt {{t^2} + 1} + \dfrac{1}{2}\ln |t + \sqrt {{t^2} + 1} |\} \\
   = 2\{ \tan y.\sec y + \ln |\tan y + \sec y|\} \;
  $
 $ (\because t = \tan y) $
Now, returning to:
 $ \sec y = \dfrac{x}{2} $ , so that: $ \tan y = \sqrt {\dfrac{{{x^2}}}{4} - 1} $
SO, we have now:
 $
  \therefore I = 2\{ \dfrac{x}{2}.\dfrac{{\sqrt {{x^2} - 4} }}{2} + \ln |\dfrac{x}{2} + \dfrac{{\sqrt {{x^2} - 4} }}{2}|\} \\
   \Rightarrow I = \dfrac{x}{2}.\sqrt {{x^2} - 4} + 2\ln |\dfrac{x}{2} + \dfrac{{\sqrt {{x^2} - 4} }}{2}| + C \;
   $
,or,

 $ I = \dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + c $ , $ c = C - 2\ln 2 $
However, the integral can easily be dealt with without using the substitution as follows:
 $
  I = \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx \\
   = \int \{ \dfrac{{({x^2} - 4) + 4}}{{\sqrt {{x^2} - 4} }}dx\, \\
   = \int \{ \dfrac{{({x^2} - 4)}}{{\sqrt {{x^2} - 4} }} + \dfrac{4}{{\sqrt {{x^2} - 4} }}\} dx \\
   = \int \sqrt {{x^2} - 4} dx + 4\int \dfrac{1}{{\sqrt {{x^2} - 4} }}dx \\
   = \{ \dfrac{x}{2}\sqrt {{x^2} - 4} - \dfrac{4}{2}\ln |x + \sqrt {{x^2} - 4} |\} + 4\ln |x + \sqrt {{x^2} - 4} | \;
  $
 $ \therefore I = \dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + {C_1} $
So, the correct answer is “ $ \dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + {C_1} $ ”.

Note: Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.