# Antiderivative of arccos

## Theorem

The following formula holds: $$\displaystyle\int \mathrm{arccos}(z) \mathrm{d}z = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C,$$ where $\mathrm{arccos}$ denotes the inverse cosine function and $C$ denotes an arbitrary constant.

## Proof

Let $u=\mathrm{arccos}(z)$ so by derivative of arccos we know that $\mathrm{d}u=-\dfrac{1}{\sqrt{1-z^2}} \mathrm{d}z$ and let $\mathrm{d}v=1 \mathrm{d}z$ so that $v=z$. Then, integration by parts yields $$\begin{array}{ll} \displaystyle\int \mathrm{arccos}(z) \mathrm{d}z &=z\mathrm{arccos}(z) + \displaystyle\int \dfrac{z}{\sqrt{1-z^2}} \mathrm{d}z \\ &\stackrel{w=1-z^2}{=} z\mathrm{arccos}(z)-\dfrac{1}{2} \displaystyle\int w^{-\frac{1}{2}} \mathrm{d}w \\ &= z\mathrm{arccos}(z) - \sqrt{w} + C \\ &= z\mathrm{arccos}(z) - \sqrt{1-z^2} + C, \end{array}$$ as was to be shown.