Circular slice
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Problem

Use the substitution \(x = \cos \theta\) to find

\[\int_a^b \sqrt{1-x^2} \, dx\]

If \(x = \cos \theta\) then \(\dfrac{dx}{d\theta} = -\sin \theta\), meaning we can write \(dx = -\sin \theta \, d\theta.\) \[\begin{align*} \int_a^b \sqrt{1-x^2} \, dx &= \int\limits_{\arccos a}^{\arccos b} \!\!\!\! \sqrt{1-\cos^2 \theta} \times -\sin \theta \, d\theta \\ &= \int\limits_{\arccos a}^{\arccos b} \!\!\!\! -\sin^2 \theta \, d\theta \\ \end{align*}\]

We can’t integrate \(\sin^2 \theta\) directly, instead we need to recognise that we can rewrite it in terms of \(\cos 2\theta\).

You might have noticed that the limits have changed. Why is this the case?

\[\begin{align*} \int\limits_{\arccos a}^{\arccos b} \!\!\!\! - \sin^2 \theta \, d\theta &= \int\limits_{\arccos a}^{\arccos b} \!\!\!\! \left(\frac{1}{2} \cos 2\theta - \frac{1}{2}\right) \, d\theta \\ &= \left[\frac{1}{4}\sin 2\theta - \frac{1}{2}\theta \right]_{\arccos a}^{\arccos b} \\ &= \left[\frac{1}{2}\sin\theta \cos \theta - \frac{1}{2}\theta \right]_{\arccos a}^{\arccos b} \end{align*}\]

From here we could substitute in our values of \(\theta\) but we will get expressions like \(\sin(\arccos b)\) which can be difficult to work with. Instead it will be helpful to rewrite our expression in terms of \(x\).

The functions \(\sin \theta\), \(\cos \theta\) and \(\theta\) appear. We know \(\cos \theta = x\), as this is our original substitution, and therefore \(\theta = \arccos x\). For \(\sin \theta\) we can think of \(\sin^2\theta + \cos^2 \theta = 1\), or a right angle triangle with a hypotenuse of \(1\) and base of \(x\), both of which give us the same result: \(\sin \theta = \sqrt{1-x^2}.\)

\[\begin{align*} &= \left[\frac{1}{2}x\sqrt{1-x^2} - \frac{1}{2}\arccos x \right]_{x=a}^{x=b} \\ &= \frac{1}{2}b\sqrt{1-b^2} - \frac{1}{2}\arccos b - \frac{1}{2}a\sqrt{1-a^2} + \frac{1}{2}\arccos a \end{align*}\]

How does your answer to \(\displaystyle{\int_a^b \sqrt{1-x^2} \, dx}\) relate to the following diagram?