Question

How do you integrate $ \smallint (\dfrac{{{x^2}}}{{\sqrt {4 - {x^2}} }})dx $ ?

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:
 $ I = \smallint \dfrac{{{x^2}}}{{\sqrt {4 - {x^2}} }}dx $
Let $ x = 2\sin (\theta ) $
Now, differentiate the upper assumed equation:
 $ dx = 2\cos (\theta )d\theta $
So:
Now, put the value of $ x $ and $ dx $ :
 $
  I = \smallint \dfrac{{4\sin {{(\theta )}^2}}}{{\sqrt {4 - 4\sin {{(\theta )}^2}} }}.2\cos (\theta )d\theta \\
   = 2\smallint \dfrac{{2\sin {{(\theta )}^2}\cos (\theta )}}{{\sqrt {1 - \sin {{(\theta )}^2}} }}d\theta \;
  $
As we know, $ 1 - \sin {(\theta )^2} = \cos {(\theta )^2} $ ,
 $ I = 2\smallint 2\sin {(\theta )^2}d\theta $
 $ [\because 2\sin {(\theta )^2} = 1 - \cos (2\theta )] $
 $
  I = 2\smallint (1 - \cos (2\theta ))d\theta \\
   = 2\theta - \sin (2\theta ) \\
   = 2(\theta - \sin (\theta )\cos (\theta )) \;
  $
Finally, if $ x = 2\sin (\theta ) $ , as we assumed in beginning, then $ \theta = {\sin ^{ - 1}}(\dfrac{x}{2}) $ and $ \cos ({\sin ^{ - 1}}(x)) = \sqrt {1 - {x^2}} $
 $
  \therefore I = 2{\sin ^{ - 1}}(\dfrac{x}{2}) - x\sqrt {1 - \dfrac{{{x^2}}}{4}} \\
   = 2{\sin ^{ - 1}}(\dfrac{x}{2}) - \dfrac{x}{2}\sqrt {4 - {x^2}} + C \;
  $
here, $ C \in R $ .
Hence, the integration of $ \smallint (\dfrac{{{x^2}}}{{\sqrt {4 - {x^2}} }})dx $ is $ 2{\sin ^{ - 1}}(\dfrac{x}{2}) - \dfrac{x}{2}\sqrt {4 - {x^2}} + C $ .
So, the correct answer is “ $ \smallint (\dfrac{{{x^2}}}{{\sqrt {4 - {x^2}} }})dx $ is $ 2{\sin ^{ - 1}}(\dfrac{x}{2}) - \dfrac{x}{2}\sqrt {4 - {x^2}} + C $ ”.

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.