Editing Inverse trigonometric functions (section)

==In calculus==

==={{anchor|Derivatives}}Derivatives of inverse trigonometric functions===
{{Main|Differentiation of trigonometric functions}}

The [[derivative]]s for complex values of ''z'' are as follows:
:<math>\begin{align}
\frac{d}{dz} \arcsin(z) &{} = \frac{1}{\sqrt{1-z^2}} \; ;   &z &{}\neq -1, +1 \\
\frac{d}{dz} \arccos(z) &{} = -\frac{1}{\sqrt{1-z^2}} \; ;  &z &{}\neq -1, +1 \\
\frac{d}{dz} \arctan(z) &{} = \frac{1}{1+z^2} \; ;          &z &{}\neq -i, +i\\
\frac{d}{dz} \arccot(z) &{} = -\frac{1}{1+z^2} \; ;         &z &{}\neq -i, +i \\
\frac{d}{dz} \arcsec(z) &{} = \frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ;   &z &{}\neq -1, 0, +1 \\
\frac{d}{dz} \arccsc(z) &{} = -\frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ;  &z &{}\neq -1, 0, +1
\end{align}</math>
Only for real values of ''x'':
:<math>\begin{align}
\frac{d}{dx} \arcsec(x) &{} = \frac{1}{|x| \sqrt{x^2-1}} \; ;  & |x| > 1\\
\frac{d}{dx} \arccsc(x) &{} = -\frac{1}{|x| \sqrt{x^2-1}} \; ; & |x| > 1
\end{align}</math>

These formulas can be derived in terms of the derivatives of trigonometric functions. For example, if <math>x = \sin \theta</math>, then <math display=inline>dx/d\theta = \cos \theta = \sqrt{1-x^2},</math> so
:<math>\frac{d}{dx}\arcsin(x) = \frac{d \theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{\sqrt{1-x^2}}.</math>

===Expression as definite integrals===
Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:
:<math>\begin{align}
\arcsin(x) &{}= \int_0^x \frac{1}{\sqrt{1 - z^2}} \, dz \; , & |x| &{} \leq 1\\
\arccos(x) &{}= \int_x^1 \frac{1}{\sqrt{1 - z^2}} \, dz \; , & |x| &{} \leq 1\\
\arctan(x) &{}= \int_0^x \frac{1}{z^2 + 1} \, dz \; ,\\
\arccot(x) &{}= \int_x^\infty \frac{1}{z^2 + 1} \, dz \; ,\\
\arcsec(x) &{}= \int_1^x \frac{1}{z \sqrt{z^2 - 1}} \, dz = \pi + \int_{-x}^{-1} \frac{1}{z \sqrt{z^2 - 1}} \, dz\; , & x &{} \geq 1\\
\arccsc(x) &{}= \int_x^\infty \frac{1}{z \sqrt{z^2 - 1}} \, dz = \int_{-\infty}^{-x} \frac{1}{z \sqrt{z^2 - 1}} \, dz \; , & x &{} \geq 1\\
\end{align}</math>
When ''x'' equals 1, the integrals with limited domains are [[improper integral]]s, but still well-defined.

===Infinite series===
Similar to the sine and cosine functions, the inverse trigonometric functions can also be calculated using [[power series]], as follows. For arcsine, the series can be derived by expanding its derivative, <math display="inline">\tfrac{1}{\sqrt{1-z^2}}</math>, as a [[binomial series]], and integrating term by term (using the integral definition as above). The series for arctangent can similarly be derived by expanding its derivative <math display="inline">\frac{1}{1+z^2}</math> in a [[geometric series]], and applying the integral definition above (see [[Leibniz series]]).

: <math>
\begin{align}
\arcsin(z) & = z + \left( \frac{1}{2} \right) \frac{z^3}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \frac{z^5}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \frac{z^7}{7} + \cdots \\[5pt]
& = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{z^{2n+1}}{2n+1} \\[5pt]
& = \sum_{n=0}^\infty \frac{(2n)!}{(2^n n!)^2} \frac{z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1
\end{align}
</math>

:<math>\arctan(z)
= z - \frac{z^3}{3} +\frac{z^5}{5} - \frac{z^7}{7} + \cdots
= \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1 \qquad z \neq i,-i</math>

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above.  For example, <math>\arccos(x) = \pi/2 - \arcsin(x)</math>, <math>\arccsc(x) = \arcsin(1/x)</math>, and so on.  Another series is given by:<ref name="Borwein_2004"/>

:<math>2\left(\arcsin\left(\frac{x}{2}\right) \right)^2 = \sum_{n=1}^\infty \frac{x^{2n}}{n^2\binom {2n} n}.</math>

[[Leonhard Euler]] found a series for the arctangent that converges more quickly than its [[Taylor series]]:

: <math>\arctan(z) = \frac z {1 + z^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k z^2}{(2k + 1)(1 + z^2)}.</math><ref>{{citation|title= An elementary derivation of Euler's series for the arctangent function|journal = The Mathematical Gazette | author = Hwang Chien-Lih | doi = 10.1017/S0025557200178404 | year = 2005 | volume = 89 | issue = 516|pages = 469–470 |s2cid = 123395287 }}</ref>
(The term in the sum for ''n'' = 0 is the [[empty product]], so is 1.)

Alternatively, this can be expressed as

:<math>\arctan(z) = \sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n + 1)!} \frac{z^{2n + 1}}{(1 + z^2)^{n + 1}}.</math>

Another series for the arctangent function is given by

:<math>\arctan(z) = i\sum_{n=1}^\infty\frac{1}{2n - 1}\left(\frac{1}{(1 + 2i/z)^{2n-1}} - \frac{1}{(1 - 2i/z)^{2n - 1}}\right),</math>

where <math>i=\sqrt{-1}</math> is the [[imaginary unit]].<ref>{{citation | title = A formula for pi involving nested radicals | journal = The Ramanujan Journal | author = S. M. Abrarov and B. M. Quine | doi = 10.1007/s11139-018-9996-8 | year = 2018 | volume = 46 | issue = 3 | pages = 657–665 | arxiv = 1610.07713 | s2cid = 119150623 }}</ref>

====Continued fractions for arctangent====
Two alternatives to the power series for arctangent are these [[generalized continued fraction]]s:

: <math>\arctan(z) =
\frac z {1 + \cfrac{(1z)^2}{3 - 1z^2 + \cfrac{(3z)^2}{5 - 3z^2 + \cfrac{(5z)^2}{7 - 5z^2 + \cfrac{(7z)^2}{9-7z^2 + \ddots}}}}} =
\frac{z}{1 + \cfrac{(1z)^2}{3 + \cfrac{(2z)^2}{5 + \cfrac{(3z)^2}{7 + \cfrac{(4z)^2}{9 + \ddots}}}}}
</math>

The second of these is valid in the cut complex plane. There are two cuts, from &minus;'''i''' to the point at infinity, going down the imaginary axis, and from '''i''' to the point at infinity, going up the same axis. It works best for real numbers running from −1 to 1. The partial denominators are the odd natural numbers, and the partial numerators (after the first) are just (''nz'')<sup>2</sup>, with each perfect square appearing once. The first was developed by [[Leonhard Euler]]; the second by [[Carl Friedrich Gauss]] utilizing the [[Gaussian hypergeometric series]].

===Indefinite integrals of inverse trigonometric functions===

For real and complex values of ''z'':
:<math>\begin{align}
\int \arcsin(z) \, dz &{}= z \, \arcsin(z) + \sqrt{1 - z^2} + C\\
\int \arccos(z) \, dz &{}= z \, \arccos(z) - \sqrt{1 - z^2} + C\\
\int \arctan(z) \, dz &{}= z \, \arctan(z) - \frac{1}{2} \ln \left( 1 + z^2 \right) + C\\
\int \arccot(z) \, dz &{}= z \, \arccot(z) + \frac{1}{2} \ln \left( 1 + z^2 \right) + C\\
\int \arcsec(z) \, dz &{}= z \, \arcsec(z) - \ln \left[ z \left( 1 + \sqrt{ \frac{z^2-1}{z^2} } \right) \right] + C\\
\int \arccsc(z) \, dz &{}= z \, \arccsc(z) + \ln \left[ z \left( 1 + \sqrt{ \frac{z^2-1}{z^2} } \right) \right] + C
\end{align}</math>

For real ''x'' ≥ 1:
:<math>\begin{align}
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \ln \left( x + \sqrt{x^2-1} \right) + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \ln \left( x + \sqrt{x^2-1} \right) + C
\end{align}</math>

For all real ''x'' not between -1 and 1:
:<math>\begin{align}
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \sgn(x) \ln\left| x + \sqrt{x^2-1}\right| + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \sgn(x) \ln\left| x + \sqrt{x^2-1}\right| + C
\end{align}</math>

The absolute value is necessary to compensate for both negative and positive values of the arcsecant and arccosecant functions. The signum function is also necessary due to the absolute values in the [[#Derivatives|derivative]]s of the two functions, which create two different solutions for positive and negative values of x. These can be further simplified using the logarithmic definitions of the [[Inverse hyperbolic functions#Logarithmic representation|inverse hyperbolic function]]s:
:<math>\begin{align}
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \operatorname{arcosh}(|x|) + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \operatorname{arcosh}(|x|) + C\\
\end{align}</math>

The absolute value in the argument of the arcosh function creates a negative half of its graph, making it identical to the signum logarithmic function shown above.

All of these antiderivatives can be derived using [[integration by parts]] and the simple derivative forms shown above.

====Example====
Using <math>\int u \, dv = u v - \int v \, du</math> (i.e. [[integration by parts]]), set

:<math>\begin{align}
u &= \arcsin(x) & dv &= dx \\
du &= \frac{dx}{\sqrt{1-x^2}} & v &= x
\end{align}</math>

Then

:<math>\int \arcsin(x) \, dx = x \arcsin(x) - \int \frac{x}{\sqrt{1-x^2}} \, dx,</math>

which by the simple [[Integration by substitution|substitution]] <math>w=1-x^2,\ dw = -2x\,dx</math> yields the final result:

:<math>\int \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1-x^2} + C </math>