Two ways of doing an integral (Part 2)

A colleague placed the following problem on an exam, expecting the following solution:

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

However, one student produced the following solution (see yesterday’s post for details):

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

Here’s how I showed they are equivalent.

Let \alpha = \displaystyle \sin^{-1} \left( \frac{x-2}{2} \right) and \beta = \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) . Then

\displaystyle \sin(\alpha - 2\beta) = \sin \alpha \cos 2\beta - \cos \alpha \sin 2\beta.

Let’s evaluate the four expressions on the right-hand side.

First, \sin \alpha is clearly equal to \displaystyle \frac{x-2}{2}.

Second, \cos 2\beta = 1 - 2 \sin^2 \beta, so that

\cos 2\beta = \displaystyle 1 - 2\left( \frac{\sqrt{x}}{2} \right)^2 = \displaystyle 1 - \frac{x}{2} = \displaystyle -\frac{x-2}{2}.

Third, to evaluate $\cos \alpha$, I’ll use the identity \displaystyle \cos \left( \sin^{-1} x \right) = \sqrt{1 - x^2}:

\cos \alpha = \displaystyle \sqrt{1 - \left( \frac{x-2}{2} \right)^2 } = \displaystyle \frac{\sqrt{4x-x^2}}{2}

Fourth, \sin 2\beta = 2 \sin \beta \cos \beta. Using the above identity again, we find

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

= \sqrt{x} \sqrt{1 - \displaystyle \frac{x}{4}}

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

 Combining the above, we find

\sin(\alpha - 2 \beta) = \displaystyle -\left( \frac{x-2}{2} \right)^2 - \left( \frac{\sqrt{4x-x^2}}{2} \right)^2

\sin(\alpha - 2 \beta) = \displaystyle \frac{-(x^2 - 4x + 4) - (4x - x^2)}{4}

\sin(\alpha - 2 \beta) = -1

\alpha - 2 \beta = \displaystyle -\frac{\pi}{2} + 2\pi n for some integer n

Also, since -\pi/2 \le \alpha \le \pi/2 and 0 \le -2\beta \le \pi, we see that -\pi/2 \le \alpha - 2 \beta \le 3\pi/2. (From its definition, $\beta$ is the arcsine of a positive number and therefore must be nonnegative.) Therefore, \alpha - 2\beta = -\pi/2.

In other words,

\sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) and 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)

differ by a constant, thus showing that the two antiderivatives are equivalent.


Leave a comment

1 Comment

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

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: