A calculus of the absurd

12.5.3 Integral of \(\frac {1-\cos (x)}{1+\cos (x)}\)
  • Example 23 Find the value of

    \begin{equation*} \int \frac {1-\cos (x)}{1+\cos (x)} dx \end{equation*}

Solution: Anything which looks like \(1+\cos (x)\) can be turned into \(1-\cos ^2(x)\) which is also known9090 By the Pythagorean identity. as \(\sin ^2(x)\). This is by multiplying by \(1-\cos (x)\) (because \((x+y)(x-y)=x^2-y^2\), also known as the difference of two squares). We can proceed by multiplying by one.

\begin{align*} \int \frac {1-\cos (x)}{1+\cos (x)} dx &= \int \frac {1-\cos (x)}{1+\cos (x)} \cdot \frac {1-\cos (x)}{1-\cos (x)} dx \\ &= \int \frac {1 - 2\cos (x) + \cos ^2(x)}{\sin ^2(x)} dx \\ &= \int \csc ^2(x) - 2\cot (x)\csc (x) + \cot ^2(x) dx \end{align*}

Note that while a lot of these integrals are in the formula sheet 9191 At least, in the OCR A formula booklet, it is given that \(\frac {d}{dx}\left [ \cot (x) \right ] = - \csc ^2(x) \) and \(\frac {d}{dx}\left [ \csc (x) \right ] = - \csc (x)\cot (x) \) some of them are not (i.e. we have no idea what the integral of \(\cot ^2(x)\) is at present). To get around this, we can use the Pythagorean identity to reduce this problem (the answer to which we don’t know) to a problem which we do know the answer to! Take \(\cos ^2(x) + \sin ^2(x) = 1\), divide through by \(\sin ^2(x)\) and obtain that \(\cot ^2(x) + 1 = \csc ^2(x) \implies \cot ^2(x) = \csc ^2(x) - 1\).

Then,

\begin{align*} \int \csc ^2(x) - 2\cot (x)\csc (x) + \cot ^2(x) dx &= \int 2\csc ^2(x) - 2\cot (x)\csc (x) - 1 \\ &= -2\cot ^2(x) + 2\csc (x) - x + c \end{align*}