# A calculus of the absurd

##### 11.5.3 Integral of $$\frac {1-\cos (x)}{1+\cos (x)}$$
• Example 11.5.3 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 known8686 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 8787 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,