Which substitution?

Excerpt: Students are asked to do the same integral twice, using a different substitution each time. The problem highlights that you can arrive at different answers because of this, and asks students to think about how the two answers are related.

Problem

$\displaystyle{\int \dfrac{1}{\sqrt{1-x^2}} \, dx}$ can be integrated by using a substitution. What answer do you get if you use

$x = \sin \theta$ ?

$x = \cos \theta$ ?

Do they both give the correct answer? If so, why?

Solution

If $x = \sin \theta$ , then $\dfrac{dx}{d\theta} = \cos \theta$ , meaning we can write $dx = \cos \theta \ d\theta$ . Substituting this into the integral we get

\begin{align*} \int \dfrac{1}{\sqrt{1-x^2}} \, dx &= \int \dfrac{1}{\sqrt{1-\sin^2 \theta}} \ \cos \theta \, d\theta\\ &= \int \dfrac{1}{\cos \theta} \ \cos \theta \, d\theta \\ &= \int 1 \, d\theta \\ &= \theta + c. \end{align*}

As $x = \sin \theta$ , then $\theta = \arcsin x$ , so the answer in terms of $x$ is $\displaystyle{\int \dfrac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + c}.$

Remember that inverse trigonometric functions can be written in two different ways. For example $\arcsin$ and $\sin^{-1}$ .

This time $x = \cos \theta$ , so $\dfrac{dx}{d\theta} = -\sin \theta$ . Writing $dx = -\sin \theta \ d\theta$ gives us

\begin{align*} \int \dfrac{1}{\sqrt{1-x^2}} \, dx &= \int \dfrac{1}{\sqrt{1-\cos^2 \theta}} \cdot (-\sin \theta) \, d\theta\\ &= \int -1 \, d\theta \\ &= -\theta + c. \end{align*}

As $x = \cos \theta$ , $\theta = \arccos x$ , so this time we get the answer $\displaystyle{\int \dfrac{1}{\sqrt{1-x^2}} \, dx = -\arccos x + c}.$

The fact that the answers are both inverse trigonometric functions suggests there is a relationship between them. The easiest way to think about their differences and similarities is to sketch the graphs of both functions. We see that

y = -\arccos x

is the same shape as

y = \arcsin x

, the only difference being a translation of

\frac{π}{2}

along the

y

axis. This difference can be accounted for by the constants of integration.

We could also think about this algebraically. If we let $y = \arcsin x$ , then $\sin y = x.$ As $\sin y = \cos(\frac{\pi}{2} - y)$ we can write

\begin{align*} &&\cos\left(\frac{\pi}{2} - y\right) &= x \\ \iff&& \frac{\pi}{2} - y &= \arccos x \\ \iff&& \frac{\pi}{2} - \arcsin x &= \arccos x. \end{align*}

The second line of this argument will not always follow on from the first, as it depends on the range of values for $y$ . Can you see why it is valid in this case?

This method also shows that the values of $\arcsin x$ and $-\arccos x$ differ by $\frac{\pi}{2},$ or in fact, that $\arcsin x + \arccos x = \frac{\pi}{2}.$

Reflection

Does having a different antiderivative function affect the solution to a definite integral? Remember, $\int_a^b f(x) \, dx = F(b) - F(a).$

It is useful to realise that there can be more than one form of answer to an integral. This happens particularly when using trigonometric substitutions as there are so many ways that different functions can be written in terms of each other.