Two ways of doing an integral (Part 1)

A colleague placed the following problem on an exam:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}}$

He expected students to solve this problem by the standard technique, completing the square:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \displaystyle \int \frac{dx}{4-(x-2)^2} = \sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) + C$

However, one student solved this problem by some clever algebra and the substitution $u = \sqrt{x}$, so that $x = u^2$ and $dx = 2u \, du = 2 \sqrt{x} \, du$:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \displaystyle \int \frac{dx}{\sqrt{x} \sqrt{4-x}}$

$= \displaystyle \int \frac{dx}{\sqrt{x} \sqrt{4 - (\sqrt{x})^2}}$

$= \displaystyle \int \frac{2 \, du}{\sqrt{4 - u^2}}$

$= 2 \sin^{-1} \left( \displaystyle \frac{u}{2} \right) + C$

$= 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) + C$

As he couldn’t find a mistake in the student’s work, he assumed that the two expressions were equivalent. Indeed, he differentiated the student’s work to make sure it was right. But he couldn’t immediately see, using elementary reasoning, why they were equivalent. So he walked across the hall to my office to ask me if I could help.

After a few minutes, I was able to show that the two expressions were equivalent.

I’ll leave this one as a cliff-hanger for now. In tomorrow’s post, I’ll show why they’re equivalent.