Difference between revisions of "Antiderivative of arccos"
From specialfunctionswiki
(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\displaystyle\int \mathrm{arccos}(z...") |
|||
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\displaystyle\int \mathrm{arccos}(z) \mathrm{d}z = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C,$$ | $$\displaystyle\int \mathrm{arccos}(z) \mathrm{d}z = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C,$$ | ||
where $\mathrm{arccos}$ denotes the [[arccos|inverse cosine]] function and $C$ denotes an arbitrary constant. | where $\mathrm{arccos}$ denotes the [[arccos|inverse cosine]] function and $C$ denotes an arbitrary constant. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
Revision as of 07:24, 8 June 2016
Theorem
The following formula holds: $$\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.
