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.