Editing Inverse trigonometric functions (section)

===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>