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

  1. \(x = \sin \theta\)?
  2. \(x = \cos \theta\)?
  1. 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 \(\sin^{-1}\).

  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 \[\displaystyle{\int \dfrac{1}{\sqrt{1-x^2}} \, dx = -\arccos x + c}.\]

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

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 \(\arcsin x + \arccos x = \frac{\pi}{2}.\)


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.