Difference between revisions of "Antiderivative of arccos"
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| − | + | ==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== | |
| − | + | 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. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 21:15, 28 March 2017
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.
