Inverse Trigonometric Functions

In mathematics, the inverse trigonometric functions (occasionally also called arcus functions, antitrigonometric functions or cyclometric functions are the inverse functions of the trigonometric functions (with suitably restricted domains). Specifically, they are the inverses of the sine, cosine, tangent (trigonometry)|tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.

Notation

Several notations for the inverse trigonometric functions exist. The most common convention is to name inverse trigonometric functions using an arc- prefix: arcsin(x), arccos(x), arctan(x), etc. This notation arises from the following geometric relationships: when measuring in radians, an angle of θ radians will correspond to an arc whose length is rθ, where r is the radius of the circle. Thus in the unit circle, "the arc whose cosine is x" is the same as "the angle whose cosine is x", because the length of the arc of the circle in radii is the same as the measurement of the angle in radians. In computer programming languages, the inverse trigonometric functions are often called by the abbreviated forms asin, acos, atan.

The notations sin⁻¹(x), cos⁻¹(x), tan⁻¹(x), etc., as introduced by John Herschel in 1813, are often used as well in English-language sources—conventions consistent with the notation of an inverse function. This might appear to conflict logically with the common semantics for expressions such as sin²(x), which refer to numeric power rather than function composition, and therefore may result in confusion between multiplicative inverse or reciprocal and compositional inverse. The confusion is somewhat mitigated by the fact that each of the reciprocal trigonometric functions has its own name—for example, (cos(x))⁻¹ = sec(x). Nevertheless, certain authors advise against using it for its ambiguity. Another convention used by a few authors is to use an uppercase first letter, along with a ⁻¹ superscript: Sin⁻¹(x), Cos⁻¹(x), Tan⁻¹(x), etc. This potentially avoids confusion with the multiplicative inverse, which should be represented by sin⁻¹(x), cos⁻¹(x), etc.

Since 2009, the ISO 80000-2 standard has specified solely the "arc" prefix for the inverse functions.

Basic concepts

Principal values

Since none of the six trigonometric functions are one-to-one, they must be restricted in order to have inverse functions. Therefore, the result ranges of the inverse functions are proper subsets of the domains of the original functions.

For example, using function in the sense of multivalued functions, just as the square root function $y={\sqrt {x}}$ could be defined from $y^{2}=x,$ the function $y=\arcsin(x)$ is defined so that $\sin(y)=x.$ For a given real number $x,$ with $-1\leq x\leq 1,$ there are multiple (in fact, countably infinite) numbers $y$ such that $\sin(y)=x$; for example, $\sin(0)=0,$ but also $\sin(\pi )=0,$ $\sin(2\pi )=0,$ etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each $x$ in the domain, the expression $\arcsin(x)$ will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.

The principal inverses are listed in the following table.

Name	Usual notation	Definition	Domain of $x$ for real result	Range of usual principal value (radians)	Range of usual principal value (degrees)
arcsine	$y=\arcsin(x)$	x = sin(y)	$-1\leq x\leq 1$	$-{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}$	$-90^{\circ }\leq y\leq 90^{\circ }$
arccosine	$y=\arccos(x)$	x = cos(y)	$-1\leq x\leq 1$	$0\leq y\leq \pi$	$0^{\circ }\leq y\leq 180^{\circ }$
arctangent	$y=\arctan(x)$	x = tan(y)	all real numbers	$-{\frac {\pi }{2}}<y<{\frac {\pi }{2}}$	$-90^{\circ }<y<90^{\circ }$
arccotangent	$y=\operatorname {arccot}(x)$	x = cot(y)	all real numbers	$0<y<\pi$	$0^{\circ }<y<180^{\circ }$
arcsecant	$y=\operatorname {arcsec}(x)$	x = sec(y)	$x\leq -1{\text{ or }}x\geq 1$	$0\leq y<{\frac {\pi }{2}}{\text{ or }}{\frac {\pi }{2}}<y\leq \pi$	$0^{\circ }\leq y<90^{\circ }{\text{ or }}90^{\circ }<y\leq 180^{\circ }$
arccosecant	$y=\operatorname {arccsc}(x)$	x = csc(y)	$x\leq -1{\text{ or }}x\geq 1$	$-{\frac {\pi }{2}}\leq y<0{\text{ or }}0<y\leq {\frac {\pi }{2}}$	$-90^{\circ }\leq y<0^{\circ }{\text{ or }}0^{\circ }<y\leq 90^{\circ }$

Note: Some authors define the range of arcsecant to be ${\big (}0\leq y<{\tfrac {\pi }{2}}{\text{ or }}\pi <y\leq {\tfrac {3\pi }{2}}{\big )}$, because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, $\tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}},$ whereas with the range ${\big (}0\leq y<{\tfrac {\pi }{2}}{\text{ or }}{\tfrac {\pi }{2}}<y\leq \pi {\big )}$, we would have to write $\tan(\operatorname {arcsec}(x))=\pm {\sqrt {x^{2}-1}},$ since tangent is nonnegative on $0\leq y<{\tfrac {\pi }{2}},$ but nonpositive on ${\tfrac {\pi }{2}}<y\leq \pi .$ For a similar reason, the same authors define the range of arccosecant to be ${\big (}-\pi <y\leq -{\tfrac {\pi }{2}}{\text{ or }}0<y\leq {\tfrac {\pi }{2}}{\big )}.$

If $x$ is allowed to be a complex number, then the range of $y$ applies only to its real part.

Solutions to elementary trigonometric equations

Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $2\pi$:

• Sine and cosecant begin their period at $2\pi k-{\frac {\pi }{2}}$ (where $k$ is an integer), finish it at $2\pi k+{\frac {\pi }{2}},$ and then reverse themselves over $2\pi k+{\frac {\pi }{2}},$ to $2\pi k+{\frac {3\pi }{2}}.$
• Cosine and secant begin their period at $2\pi k,$ finish it at $2\pi k+\pi .$ and then reverse themselves over $2\pi k+\pi$ to $2\pi k+2\pi .$
• Tangent begins its period at $2\pi k-{\frac {\pi }{2}},$ finishes it at $2\pi k+{\frac {\pi }{2}},$ and then repeats it (forward) over $2\pi k+{\frac {\pi }{2}}$ to $2\pi k+{\frac {3\pi }{2}}.$
• Cotangent begins its period at $2\pi k,$ finishes it at $2\pi k+\pi ,$ and then repeats it (forward) over $2\pi k+\pi$ to $2\pi k+2\pi .$

This periodicity is reflected in the general inverses, where $k$ is some integer.

For example, if $\cos \theta =-1$ then $\theta =\pi +2\pi k=-\pi +2\pi (1+k)$ for some $k\in \mathbb {Z} .$ While if $\sin \theta =\pm 1$ then $\theta ={\frac {\pi }{2}}+\pi k=-{\frac {\pi }{2}}+\pi (k+1)$ for some $k\in \mathbb {Z} ,$ where $k$ will be even if $\sin \theta =1$ and it will be odd if $\sin \theta =-1.$ The equations $\sec \theta =-1$ and $\csc \theta =\pm 1$ have the same solutions as $\cos \theta =-1$ and $\sin \theta =\pm 1,$ respectively. In all equations above except for those just solved (i.e. except for $\sin$/$\csc \theta =\pm 1$ and $\cos$/$\sec \theta =-1$), the integer $k$ in the solution's formula is uniquely determined by $\theta$ (for fixed $r,s,x,$ and $y$).

Detailed example and explanation of the "plus or minus" symbol $\pm$

The solutions to $\cos \theta =x$ and $\sec \theta =x$ involve the "plus or minus" symbol $\,\pm ,\,$ whose meaning is now clarified. Only the solution to $\cos \theta =x$ will be discussed since the discussion for $\sec \theta =x$ is the same. We are given $x$ between $-1\leq x\leq 1$ and we know that there is an angle $\theta$ in some give interval that satisfies $\cos \theta =x.$ We want to find this $\theta .$ The formula for the solution involves:

$$\pm \arccos x.$$

If $\,\arccos x=0\,$ (which only happens when $x=1$) then $\,+\arccos x=0\,$ and $\,-\arccos x=0\,$ so either way, $\,\pm \arccos x\,$ can only be equal to $0.$ But if $\,\arccos x\neq 0,\,$ which will now be assumed, then the solution to $\cos \theta =x,$ which is written above as

$$\,\theta =\pm \arccos x+2\pi k\,\quad {\text{ for some }}k\in \mathbb {Z}$$

is shorthand for the following statement: Either

• $\,\theta =\arccos x+2\pi k\,$ for some integer $k,$
  or else
• $\,\theta =-\arccos x+2\pi k\,$ for some integer $k.$

Because $\,\arccos x\neq 0\,$ and $\,0<\arccos x\leq \pi \,$ exactly one of these two equalities can hold. Additional information about $\theta$ is needed to determine which one holds. For example, suppose that $x=0$ and that all that is known about $\theta$ is that $\,-\pi \leq \theta \leq \pi \,$ (and nothing more is known). Then

$$\arccos x=\arccos 0={\frac {\pi }{2}}$$

and moreover, in this particular case $k=0$ (for both the $\,+\,$ case and the $\,-\,$ case) and so consequently,

$$\theta ~=~\pm \arccos x+2\pi k~=~\pm \left({\frac {\pi }{2}}\right)+2\pi (0)~=~\pm {\frac {\pi }{2}}.$$

This means that $\theta$ could be either $\,\pi /2\,$ or $\,-\pi /2.$ Without additional information it is not possible to determine which of these values $\theta$ has. An example of some additional information that could determine the value of $\theta$ would be knowing that the angle is above the $x$-axis (in which case $\theta =\pi /2$) or alternatively, knowing that it is below the $x$-axis (in which case $\theta =-\pi /2$).

Transforming equations

The equations above can be transformed by using the identities

$\operatorname {func} \$		$\operatorname {func} (-\theta )$	$\operatorname {func} \left({\frac {\pi }{2}}-\theta \right)$	$\operatorname {func} (\pi -\theta )$	$\operatorname {func} (2\pi -\theta )$
$\scriptstyle |\,\scriptstyle |$		$\scriptstyle |\,\scriptstyle |$	$\scriptstyle |\,\scriptstyle |$	$\scriptstyle |\,\scriptstyle |$	$\scriptstyle |\,\scriptstyle |$
$\sin \$		$-\sin \theta$	$\cos \theta$	$\sin \theta$	$-\sin \theta$
$\cos \$		$\cos \theta$	$\sin \theta$	$-\cos \theta$	$\cos \theta$
$\tan \$		$-\tan \theta$	$\cot \theta$	$-\tan \theta$	$-\tan \theta$
$\csc \$		$-\csc \theta$	$\sec \theta$	$\csc \theta$	$-\csc \theta$
$\sec \$		$\sec \theta$	$\csc \theta$	$-\sec \theta$	$\sec \theta$
$\cot \$		$-\cot \theta$	$\tan \theta$	$-\cot \theta$	$-\cot \theta$

So for example, by using the equality $\sin \left({\frac {\pi }{2}}-\theta \right)=\cos \theta$, the equation $\cos \theta =x$ can be transformed into $\sin \left({\frac {\pi }{2}}-\theta \right)=x,$ which allows for the solution to the equation $\;\sin \varphi =x\;$ (where $\varphi :={\frac {\pi }{2}}-\theta$) to be used; that solution being: $\varphi =(-1)^{k}\arcsin(x)+\pi k\;{\text{ for some }}k\in \mathbb {Z} ,$ which becomes:

$${\frac {\pi }{2}}-\theta ~=~(-1)^{k}\arcsin(x)+\pi k\quad {\text{ for some }}k\in \mathbb {Z}$$

where using the fact that $(-1)^{k}=(-1)^{-k}$ and substituting $h:=-k$ proves that another solution to $\;\cos \theta =x\;$ is:

$$\theta ~=~(-1)^{h+1}\arcsin(x)+\pi h+{\frac {\pi }{2}}\quad {\text{ for some }}h\in \mathbb {Z} .$$

The substitution $\;\arcsin x={\frac {\pi }{2}}-\arccos x\;$ may be used express the right hand side of the above formula in terms of $\;\arccos x\;$ instead of $\;\arcsin x.\;$

Relationships between trigonometric functions and inverse trigonometric functions

Trigonometric functions of inverse trigonometric functions are tabulated below. A quick way to derive them is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length $x,$ then applying the Pythagorean theorem and definitions of the trigonometric ratios. Purely algebraic derivations are longer. It is worth noting that for arcsecant and arccosecant, the diagram assumes that $x$ is positive, and thus the result has to be corrected through the use of absolute values and the signum (sgn) operation.

$\theta$	$\sin(\theta )$	$\cos(\theta )$	$\tan(\theta )$	Diagram
$\arcsin(x)$	$\sin(\arcsin(x))=x$	$\cos(\arcsin(x))={\sqrt {1-x^{2}}}$	$\tan(\arcsin(x))={\frac {x}{\sqrt {1-x^{2}}}}$
$\arccos(x)$	$\sin(\arccos(x))={\sqrt {1-x^{2}}}$	$\cos(\arccos(x))=x$	$\tan(\arccos(x))={\frac {\sqrt {1-x^{2}}}{x}}$
$\arctan(x)$	$\sin(\arctan(x))={\frac {x}{\sqrt {1+x^{2}}}}$	$\cos(\arctan(x))={\frac {1}{\sqrt {1+x^{2}}}}$	$\tan(\arctan(x))=x$
$\operatorname {arccot}(x)$	$\sin(\operatorname {arccot}(x))={\frac {1}{\sqrt {1+x^{2}}}}$	$\cos(\operatorname {arccot}(x))={\frac {x}{\sqrt {1+x^{2}}}}$	$\tan(\operatorname {arccot}(x))={\frac {1}{x}}$
$\operatorname {arcsec}(x)$	$\sin(\operatorname {arcsec}(x))={\frac {\sqrt {x^{2}-1}}{|x|}}$	$\cos(\operatorname {arcsec}(x))={\frac {1}{x}}$	$\tan(\operatorname {arcsec}(x))=\operatorname {sgn}(x){\sqrt {x^{2}-1}}$
$\operatorname {arccsc}(x)$	$\sin(\operatorname {arccsc}(x))={\frac {1}{x}}$	$\cos(\operatorname {arccsc}(x))={\frac {\sqrt {x^{2}-1}}{|x|}}$	$\tan(\operatorname {arccsc}(x))={\frac {\operatorname {sgn}(x)}{\sqrt {x^{2}-1}}}$

Relationships among the inverse trigonometric functions

The usual principal values of the arcsin(x) (red) and arccos(x) (blue) functions graphed on the cartesian plane.

The usual principal values of the arctan(x) and arccot(x) functions graphed on the cartesian plane.

Principal values of the arcsec(x) and arccsc(x) functions graphed on the cartesian plane.

Complementary angles:

${\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)\\[0.5em]\operatorname {arccot}(x)&={\frac {\pi }{2}}-\arctan(x)\\[0.5em]\operatorname {arccsc}(x)&={\frac {\pi }{2}}-\operatorname {arcsec}(x)\end{aligned}}$

Negative arguments:

${\begin{aligned}\arcsin(-x)&=-\arcsin(x)\\\arccos(-x)&=\pi -\arccos(x)\\\arctan(-x)&=-\arctan(x)\\\operatorname {arccot}(-x)&=\pi -\operatorname {arccot}(x)\\\operatorname {arcsec}(-x)&=\pi -\operatorname {arcsec}(x)\\\operatorname {arccsc}(-x)&=-\operatorname {arccsc}(x)\end{aligned}}$

Reciprocal arguments:

${\begin{aligned}\arccos \left({\frac {1}{x}}\right)&=\operatorname {arcsec}(x)\\[0.3em]\arcsin \left({\frac {1}{x}}\right)&=\operatorname {arccsc}(x)\\[0.3em]\arctan \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)\,,{\text{ if }}x>0\\[0.3em]\arctan \left({\frac {1}{x}}\right)&=-{\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)-\pi \,,{\text{ if }}x<0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\operatorname {arccot}(x)=\arctan(x)\,,{\text{ if }}x>0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {3\pi }{2}}-\operatorname {arccot}(x)=\pi +\arctan(x)\,,{\text{ if }}x<0\\[0.3em]\operatorname {arcsec} \left({\frac {1}{x}}\right)&=\arccos(x)\\[0.3em]\operatorname {arccsc} \left({\frac {1}{x}}\right)&=\arcsin(x)\end{aligned}}$

Useful identities if one only has a fragment of a sine table:

${\begin{aligned}\arccos(x)&=\arcsin \left({\sqrt {1-x^{2}}}\right)\,,{\text{ if }}0\leq x\leq 1{\text{ , from which you get }}\\\arccos &\left({\frac {1-x^{2}}{1+x^{2}}}\right)=\arcsin \left({\frac {2x}{1+x^{2}}}\right)\,,{\text{ if }}0\leq x\leq 1\\\arcsin &\left({\sqrt {1-x^{2}}}\right)={\frac {\pi }{2}}-\operatorname {sgn}(x)\arcsin(x)\\\arccos(x)&={\frac {1}{2}}\arccos \left(2x^{2}-1\right)\,,{\text{ if }}0\leq x\leq 1\\\arcsin(x)&={\frac {1}{2}}\arccos \left(1-2x^{2}\right)\,,{\text{ if }}0\leq x\leq 1\\\arcsin(x)&=\arctan \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\\arccos(x)&=\arctan \left({\frac {\sqrt {1-x^{2}}}{x}}\right)\\\arctan(x)&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\\\operatorname {arccot}(x)&=\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\end{aligned}}$

Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real).

A useful form that follows directly from the table above is

$\arctan \left(x\right)=\arccos \left({\sqrt {\frac {1}{1+x^{2}}}}\right)\,,{\text{ if }}x\geq 0$.

It is obtained by recognizing that $\cos \left(\arctan \left(x\right)\right)={\sqrt {\frac {1}{1+x^{2}}}}=\cos \left(\arccos \left({\sqrt {\frac {1}{1+x^{2}}}}\right)\right)$.

From the half-angle formula, $\tan \left({\tfrac {\theta }{2}}\right)={\tfrac {\sin(\theta )}{1+\cos(\theta )}}$, we get:

${\begin{aligned}\arcsin(x)&=2\arctan \left({\frac {x}{1+{\sqrt {1-x^{2}}}}}\right)\\[0.5em]\arccos(x)&=2\arctan \left({\frac {\sqrt {1-x^{2}}}{1+x}}\right)\,,{\text{ if }}-1<x\leq 1\\[0.5em]\arctan(x)&=2\arctan \left({\frac {x}{1+{\sqrt {1+x^{2}}}}}\right)\end{aligned}}$

Arctangent addition formula

$\arctan(u)\pm \arctan(v)=\arctan \left({\frac {u\pm v}{1\mp uv}}\right){\pmod {\pi }}\,,\quad uv\neq 1\,.$

This is derived from the tangent addition formula

$\tan(\alpha \pm \beta )={\frac {\tan(\alpha )\pm \tan(\beta )}{1\mp \tan(\alpha )\tan(\beta )}}\,,$

by letting

$\alpha =\arctan(u)\,,\quad \beta =\arctan(v)\,.$

In calculus

Derivatives of inverse trigonometric functions

The derivatives for complex values of z are as follows:

${\begin{aligned}{\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc}(z)&{}=-{\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\end{aligned}}$

Only for real values of x:

${\begin{aligned}{\frac {d}{dx}}\operatorname {arcsec}(x)&{}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\\{\frac {d}{dx}}\operatorname {arccsc}(x)&{}=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\end{aligned}}$

For a sample derivation: if $\theta =\arcsin(x)$, we get:

${\frac {d\arcsin(x)}{dx}}={\frac {d\theta }{d\sin(\theta )}}={\frac {d\theta }{\cos(\theta )\,d\theta }}={\frac {1}{\cos(\theta )}}={\frac {1}{\sqrt {1-\sin ^{2}(\theta )}}}={\frac {1}{\sqrt {1-x^{2}}}}$

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:

${\begin{aligned}\arcsin(x)&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz\;,&|x|&{}\leq 1\\\arccos(x)&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz\;,&|x|&{}\leq 1\\\arctan(x)&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz\;,\\\operatorname {arccot}(x)&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz\;,\\\operatorname {arcsec}(x)&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz=\pi +\int _{x}^{-1}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz\;,&x&{}\geq 1\\\operatorname {arccsc}(x)&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz=\int _{-\infty }^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz\;,&x&{}\geq 1\\\end{aligned}}$

When x equals 1, the integrals with limited domains are improper integrals, but still well-defined.

Indefinite integrals of inverse trigonometric functions

For real and complex values of z:

${\begin{aligned}\int \arcsin(z)\,dz&{}=z\,\arcsin(z)+{\sqrt {1-z^{2}}}+C\\\int \arccos(z)\,dz&{}=z\,\arccos(z)-{\sqrt {1-z^{2}}}+C\\\int \arctan(z)\,dz&{}=z\,\arctan(z)-{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arccot}(z)\,dz&{}=z\,\operatorname {arccot}(z)+{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arcsec}(z)\,dz&{}=z\,\operatorname {arcsec}(z)-\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\\\int \operatorname {arccsc}(z)\,dz&{}=z\,\operatorname {arccsc}(z)+\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\end{aligned}}$

For real x ≥ 1:

${\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}$

For all real x not between -1 and 1:

${\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\end{aligned}}$

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 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:

${\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {arcosh} (|x|)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {arcosh} (|x|)+C\\\end{aligned}}$

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 $\int u\,dv=uv-\int v\,du$ (i.e. integration by parts), set

${\begin{aligned}u&=\arcsin(x)&dv&=dx\\du&={\frac {dx}{\sqrt {1-x^{2}}}}&v&=x\end{aligned}}$

Then

$\int \arcsin(x)\,dx=x\arcsin(x)-\int {\frac {x}{\sqrt {1-x^{2}}}}\,dx,$

which by the simple substitution $w=1-x^{2},\ dw=-2x\,dx$ yields the final result:

$\int \arcsin(x)\,dx=x\arcsin(x)+{\sqrt {1-x^{2}}}+C$

Resources

Review

• Introduction to Inverse Trig Functions, Khan Academy
• Evaluating Inverse Trigonometric Functions, The Organic Chemistry Tutor

Calculus

• Derivatives of Inverse Trigonometric Functions, The Organic Chemistry Tutor

Licensing

Content obtained and/or adapted from:

• Inverse trigonometric functions, Wikipedia under a CC BY-SA license