Integrating the Integral ( int frac{x}{sqrt{4-x^4}} , dx ) Using Substitution Techniques

Integration can be a challenging but fascinating part of calculus. Consider the integral below:

The Problem

The given integral is:

[ int frac{x}{sqrt{4-x^4}} , dx ]

Step 1: Initial Substitution

To start, let's apply the substitution ( u x^2 ), which implies ( du 2x , dx ). This simplifies our integral:

[ int frac{1}{2} frac{du}{sqrt{4 - u^2}} ]

We now know the integral formula:

[ int frac{dx}{sqrt{a^2 - x^2}} sin^{-1}left( frac{x}{a} right) C ]

Applying this, we substitute ( a 2 ) and ( x u ):

[ frac{1}{2} int frac{du}{sqrt{4 - u^2}} frac{1}{2} sin^{-1}left( frac{u}{2} right) C ]

Finally, substituting back ( u x^2 ), we obtain:

[ boxed{boxed{int frac{x}{sqrt{4 - x^4}} , dx frac{1}{2} sin^{-1}left( frac{x^2}{2} right) C}} ]

Alternative Methods

Let's also explore another method, using a trigonometric substitution.

Consider ( x^2 2t ), implying ( xdx dt ). The integral transforms as follows:

[ int frac{1}{2} frac{1}{sqrt{1 - t^2}} , dt frac{1}{2} arcsin t , C ]

Substitute back ( t frac{x^2}{2} ):

[ frac{1}{2} arcsin left( frac{x^2}{2} right) , C ]

Thus, the result is the same.

Further Exploration

Another way to solve this integral, using a different substitution, is as follows:

Let ( x^2 z ), differentiating ( z ) with respect to ( x ) yields ( 2x , dx dz ).

The integral then becomes:

[ frac{1}{2} int frac{dz}{sqrt{4 - z^2}} , dz , text{Let} , u sqrt{4 - z^2} , text{implies} , du frac{-z , dz}{sqrt{4 - z^2}} ]

The integral becomes:

[ frac{1}{2} int frac{1}{u} , du frac{1}{2} sin^{-1} left( frac{z}{2} right) C ]

Substitute ( z x^2 ):

[ boxed{frac{1}{2} sin^{-1} left( frac{x^2}{2} right) C} ]

Using Additional Substitution (Trigonometric Substitution)

Lastly, consider the substitution ( x^2 2sec t ), implying ( 2x , dx 2sec t tan t , dt ).

Our integral becomes:

[ int frac{x , dx}{x^2 sqrt{x^4 - 4}} , dx , text{Let} , u sqrt{x^4 - 4} , text{implies} , du frac{2x^3 , dx}{sqrt{x^4 - 4}} ]

The integral simplifies to:

[ frac{1}{2} int frac{1}{u^2} , du -frac{1}{u} C -frac{1}{sqrt{x^4 - 4}} C ]

Substitute back to get the original variable:

[ boxed{frac{1}{4} arccos left( frac{1}{x^2} right) C} ]