Solving the Integral ( int frac{1}{x sqrt{4x^2 16}} dx ) Using Trigonometric Substitution

This article will walk you through the process of evaluating the integral ( int frac{1}{x sqrt{4x^2 16}} , dx ) using trigonometric substitution, a powerful technique in calculus. This method is especially useful for integrals involving roots of quadratic expressions.

Step 1: Simplify the Expression

First, we simplify the expression under the square root.

The given integral is:

[ int frac{1}{x sqrt{4x^2 16}} , dx ]

Notice that:

[ 4x^2 16 4x^2 4 ]

So, we can rewrite the integral as:

[ int frac{1}{x sqrt{4x^2 4}} , dx int frac{1}{x cdot 2 sqrt{x^2 1}} , dx frac{1}{2} int frac{1}{x sqrt{x^2 1}} , dx ]

Step 2: Trigonometric Substitution

To handle the term ( sqrt{x^2 1} ), we use the substitution:

[ x tan theta quad Rightarrow quad dx sec^2 theta , dtheta ]

This gives us:

[ sqrt{x^2 1} sqrt{tan^2 theta 1} sec theta ]

Step 3: Substitute into the Integral

Substituting ( x ) and ( dx ) into the integral, we get:

[ frac{1}{2} int frac{1}{tan theta cdot 2 sec theta} cdot sec^2 theta , dtheta frac{1}{2} int frac{sec theta}{2 tan theta} , dtheta frac{1}{4} int frac{sec theta}{tan theta} , dtheta ]

Recall that:

[ frac{sec theta}{tan theta} csc theta ]

So the integral simplifies to:

[ frac{1}{4} int csc theta , dtheta ]

Step 4: Integrate

The integral of ( csc theta ) is:

[ int csc theta , dtheta -ln |csc theta - cot theta| C ]

Thus:

[ frac{1}{4} int csc theta , dtheta -frac{1}{4} ln |csc theta - cot theta| C ]

Step 5: Substitute Back

Now we need to convert back to ( x ). From our substitution ( x tan theta ), we have:

[ tan theta frac{x}{1} quad cot theta frac{1}{tan theta} frac{1}{x} quad csc theta frac{1}{sin theta} sqrt{cot^2 theta 1} sqrt{1 frac{1}{x^2}} frac{sqrt{x^2 1}}{x} ]

Therefore:

[ csc theta - cot theta frac{sqrt{x^2 1}}{x} - frac{1}{x} frac{sqrt{x^2 1} - 1}{x} ]

Putting it all together, we have:

[ -frac{1}{4} ln left| frac{sqrt{x^2 1} - 1}{x} right| C ]

Final Result

The integral ( int frac{1}{x sqrt{4x^2 16}} , dx ) evaluates to:

[ -frac{1}{4} ln left| frac{sqrt{x^2 1} 1}{x} right| C ]

This solution method uses trigonometric substitution and integration by parts, providing a detailed guide on how to tackle similar integrals involving roots of quadratic expressions.