Inverse function integration formula

Statement

In terms of indefinite integrals

Suppose $f$ is a continuous one-one function. Then, we have:

$\int f^{-1}(x)dx=x{f}^{-1}(x)-\int f(u)du$

where $u=f^{-1}(x)$.

More explicitly, if $F$ is an antiderivative for $f$, then:

$\int f^{-1}(x)dx=x{f}^{-1}(x)-F(f^{-1}(x))$

This can be justified either directly or using integration by parts and integration by u-substitution.

In terms of definite integrals

Suppose $f$ is a continuous one-one function on an interval. Suppose we are integrating $f^{-1}$ on an interval of the form $[a,b]$ that lies in the range of $f$. Then:

$\underset{a}{\overset{b}{\int }}{f}^{-1}(x)=[x{f}^{-1}(x){]}_a^b-\underset{f^{-1}(a)}{\overset{f^{-1}(b)}{\int }}f(u)du$

This simplifies to:

$\underset{a}{\overset{b}{\int }}{f}^{-1}(x)=b{f}^{-1}(b)-a{f}^{-1}(a)-\underset{f^{-1}(a)}{\overset{f^{-1}(b)}{\int }}f(u)du$

More explicitly, if $F$ is an antiderivative for $f$, then:

$\underset{a}{\overset{b}{\int }}{f}^{-1}(x)=[x{f}^{-1}(x){]}_a^b-F(f^{-1}(b))+F(f^{-1}(a))$

Significance

Computational feasibility significance

Version type	Significance
indefinite integral	Given an antiderivative for a continuous one-one function $f$, it is possible to explicitly write down an antiderivative for the inverse function $f^{-1}$ in terms of $f^{-1}$ and the antiderivative for $f$.
definite integral	Given an antiderivative for a continuous one-one function $f$, and given knowledge of the values of $f^{-1}$ at $a$ and $b$, it is possible to explicitly compute the definite integral of $f^{-1}$ on $[a,b]$.

Examples

Original function	Domain on which it restricts to a one-one function	Inverse function for the restriction to that domain	Domain of inverse function (equals range of original function)	Antiderivative of original function	Antiderivative of inverse function	Explanation using inverse function integration formula	Alternate explanation using integration by parts
sine function $\sin$	$[-\pi /2,\pi /2]$	arc sine function $\arcsin$	$[-1,1]$	negative of cosine function, i.e., $-\cos$	$x\arcsin x+\sqrt{1-x^2}$	We get $x\arcsin x-(-\cos (\arcsin x))=x\arcsin x+\cos (\arcsin x)$. Now, use that $\cos$ is nonnegative on the range of $\arcsin$ and that ${\cos }^2\theta +{\sin }^2\theta =1$ to rewrite $\cos (\arcsin x)=\sqrt{1-x^2}$.	The usual method: [SHOW MORE] Take 1 as the part to integrate, get $x\arcsin x-\int \frac{xdx}{\sqrt{1-x^2}}$ For the latter integration, put $v=1-x^2$, get $(-1/2)\int \frac{dv}{\sqrt{v}}=-\sqrt{v}$. The minus sign cancels with the outer minus sign, and we get the result.
tangent function $\tan$	$(-\pi /2,\pi /2)$	arc tangent function $\arctan$	all real numbers	$-\ln \left|\cos x\right|$, same as $\ln \left|\sec x\right|$	$x\arctan x-\frac{1}{2}\ln (1+x^2)$	We get $x\arctan x-\ln |\sec (\arctan x)|$. Use that $|\sec (\arctan x)|=\sqrt{1+x^2}$ and simplify.	[SHOW MORE] Take 1 as the part to integrate, get $x\arctan x-\int \frac{xdx}{1+x^2}$, now use $g^{\prime }/g$ form to note that the latter is $(1/2)\ln (1+x^2)$.
exponential function $\exp$	all real numbers	natural logarithm $\ln$	$(0,\mathrm{\infty })$	exponential function $\exp$	$x\ln x-x$	We get $x\ln x-\exp (\ln x)=x\ln x-x$	[SHOW MORE] Take 1 as the part to integrate, get $x\ln x-\int x\frac{1}{x}dx=x\ln x-\int 1dx=x\ln x-x$.