Two ways of doing an integral (Part 1)

Calculus 1 Minute

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.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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One thought on “Two ways of doing an integral (Part 1)”

  1. Pingback: Two ways of doing an integral: Index | Mean Green Math

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