A calculus of the absurd

11.5.4 Integral of \(\sqrt {1-\cos (x)}\)

Example 11.5.4 Find the value of
\(\seteqnumber{0}{11.}{16}\)
\begin{equation} \int \sqrt {1-\cos (x)} dx \end{equation}

Solution: As it is, we can’t integrate this, therefore the only possible option is to rewrite it somehow, into a form we can integrate.

A way to think about this which isn’t particularly elegant, but is usually quite effective, is to think about all the possible identities which could work, and try each one.

e.g. thinking about identities which involve \(\cos (x)\), we have ⁸⁶⁸⁶ These are discussed in the trigonometry section of this document.

• \(\cos ^2(x) + \sin ^2(x) \equiv 1\), which doesn’t help here because there’s no \(\cos ^2(x)\) anywhere.
• \(\cos (a + b) \equiv \cos (a)\cos (b) - \sin (a)\sin (b)\) - also no help
• \(\cos (2x) \equiv \left [\cos (x)\right ]^2 - \left [\sin (x)\right ]^2\), and its two other forms (rearranging using the first bullet point), \(\cos (2x) \equiv 1 - 2\left [\sin (x)\right ]^2\) and \(\cos (2x) \equiv 2\left [\cos (x)\right ]^2 - 1\).

The last identity looks quite useful, because we have a \(\cos (2x)\), and a \(1\), and we can rewrite it in the form

\begin{equation} 1 - \cos (2x) \equiv 2\left [sin(x)\right ]^2 \end{equation}

The \(1-\cos (2x)\) looks quite a lot, but not exactly, like our integral, but we can fix that by simply replacing ⁸⁷⁸⁷ This is fine, because when we derived the double-angle formula from the addition formula we set \(a=x\) and \(b=x\) in the equation \(\cos (a + b) \equiv \cos (a)\cos (b) - \sin (a)\sin (b)\), but we could have just as well have set \(a=\frac {x}{2}\) and \(b=\frac {x}{2}\), which would give \(\cos (x) \equiv \left [\cos \left (\frac {x}{2}\right )\right ]^2 - \left [\sin \left (\frac {x}{2}\right )\right ]^2\) which we can then rearrange to give the result we need. \(x\) with \(\frac {x}{2}\), giving

\begin{equation} 1 - \cos (x) \equiv 2\left [\sin \left (\frac {x}{2}\right )\right ]^2 \end{equation}

Applying this to our integral, we have

\begin{align} \int \sqrt {1-\cos (x)} dx &= \int \sqrt { 2 \left [ \sin \left (\frac {x}{2}\right ) \right ]^2 } dx \\ &= \int \sqrt {2} \sin \left (\frac {x}{2}\right ) dx \end{align}

The last part we can do by inspection (as the derivative of \(-\cos (x)\) is equal to \(\sin (x)\), by the chain rule, the derivative of \(-\cos \left (\frac {x}{2}\right )\) is equal to \(-\frac {1}{2}\cos (\frac {x}{2})\), and therefore the integral of \(\sin \left (\frac {x}{2}\right )\) with respect to \(x\) is just \(-2\cos \left (\frac {x}{2}\right )\)).

Therefore, we have

\begin{align} \int \sqrt {1-\cos (x)} dx &= \sqrt {2}\int \sin \left (\frac {x}{2}\right ) dx \\ &= -2\sqrt {2}\cos \left (\frac {x}{2}\right ) + c \\ \end{align}