Difference between revisions of "Antiderivative of arccos"

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==Proof==
 
==Proof==
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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
 +
$$\begin{array}{ll}
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\displaystyle\int \mathrm{arccos}(z) \mathrm{d}z &=z\mathrm{arccos}(z) + \displaystyle\int \dfrac{z}{\sqrt{1-z^2}} \mathrm{d}z \\
 +
&\stackrel{w=1-z^2}{=} z\mathrm{arccos}(z)-\dfrac{1}{2} \displaystyle\int w^{-\frac{1}{2}} \mathrm{d}w \\
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&= z\mathrm{arccos}(z) - \sqrt{w} + C \\
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&= z\mathrm{arccos}(z) - \sqrt{1-z^2} + C,
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\end{array}$$
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as was to be shown.
  
 
==References==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
[[Category:Unproven]]
+
[[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.

References