Why Domain and Endpoint Checks Matter
When an inverse-trigonometric function appears in a definite integral, the interval matters as much as the integrand. The real-valued functions \(\arcsin x\) and \(\arccos x\) are defined only for \(x\) in \([-1,1]\). Their values remain finite at both endpoints, but their derivatives become unbounded there.
This does not automatically prevent integration by parts. It does mean that an endpoint calculation must be justified: the derivative formulas apply in the interior, and an interval reaching \(1\) or \(-1\) is handled by a limit from within the domain.
Real Domains, Values, and Derivatives
The principal real inverse sine and inverse cosine have the following domains, ranges, and endpoint values:
| Function | Real domain | Range | Endpoint values |
|---|---|---|---|
| \(\arcsin x\) | \([-1,1]\) | \([-\pi/2,\pi/2]\) | \(\arcsin(-1)=-\pi/2,\quad \arcsin(1)=\pi/2\) |
| \(\arccos x\) | \([-1,1]\) | \([0,\pi]\) | \(\arccos(-1)=\pi,\quad \arccos(1)=0\) |
On the open interval \((-1,1)\), their derivatives are
At \(x=-1\) and \(x=1\), the denominator is zero. These derivative formulas are therefore statements about the interior, not finite derivative values at the endpoints. The functions themselves, however, are continuous and finite on the whole closed interval.
Integration by Parts Up to a Domain Endpoint
For an interval strictly inside \((-1,1)\), the usual integration-by-parts formula applies directly. Suppose \(w\) is continuous on \([a,b]\), and let \(W\) be an antiderivative of \(w\), so \(W'=w\). Using \(u=\arcsin x\) or \(u=\arccos x\) gives the following formulas whenever the endpoints are strictly inside the domain:
The sign in the second formula is positive because the derivative of \(\arccos x\) is negative. If an endpoint is \(-1\) or \(1\), the formulas still hold when the integrals are interpreted as limits from the interior. The following argument explains why those limits exist.
Why the Endpoint Limits Are Valid
It is enough to check that the derivative terms are integrable and that the boundary terms have finite limits. The inverse-trigonometric factors are continuous and bounded on \([-1,1]\). Also, \(W\) is continuous and bounded on any closed interval \([a,b]\subseteq[-1,1]\).
For example, for \(c\) with \(-1<c<1\), the integral of the positive derivative factor from \(c\) up to \(1\), interpreted as a limit, is
This is finite. At the other endpoint, for \(-1<r<c\),
which is finite as well. Thus \(1/\sqrt{1-x^2}\) has finite integral up to either endpoint. Multiplication by the bounded function \(W\) preserves integrability. The boundary terms also have finite limits because both \(W\) and the inverse-trigonometric function have finite endpoint values.
If an endpoint is \(1\) or \(-1\), replace it temporarily by a point strictly inside \((-1,1)\).
Both inverse-trigonometric functions are differentiable on the truncated interval.
The boundary terms converge, and the derivative terms have finite improper integrals.
This is the justification for using the displayed formulas on closed intervals that touch the domain boundary. It does not assign a finite value to the derivative at the endpoint; it uses the derivative only in the interior and takes a limit.
Worked Examples
Worked Example: Integrate \(x\arcsin x\) on \([0,1]\)
Evaluate \(I=\int_0^1 x\arcsin x\,dx\). The integrand is real-valued throughout \([0,1]\). Choose \(u=\arcsin x\) and \(dv=x\,dx\). Then \(du=dx/\sqrt{1-x^2}\) in the interior and \(v=x^2/2\). Applying integration by parts with the endpoint interpreted as a limit gives
The boundary term is
To evaluate the remaining integral, put \(x=\sin t\). As \(x\) goes from \(0\) to \(1\), \(t\) goes from \(0\) to \(\pi/2\); also \(dx=\cos t\,dt\) and \(\sqrt{1-x^2}=\cos t\) on this interval. Therefore
It follows that
As a check, the direct substitution \(x=\sin t\) in the original integral gives \(I=\int_0^{\pi/2}t\sin t\cos t\,dt\). Since \(\sin t\cos t=\tfrac12\sin(2t)\), integration by parts gives
Worked Example: Integrate \(\arccos x\) on \([-1,1]\)
Evaluate \(J=\int_{-1}^1\arccos x\,dx\). The whole interval lies in the real domain. Choose \(u=\arccos x\) and \(dv=dx\), so \(du=-dx/\sqrt{1-x^2}\) in the interior and \(v=x\). Apply integration by parts first on a truncated interval and then take limits at both endpoints:
The boundary term is
The remaining integrand is odd: its numerator changes sign under \(x\mapsto -x\), while its denominator does not. Its endpoint singularities are integrable, so symmetry gives
Therefore \(J=\pi\). To check, set \(x=\cos t\), so \(t\) runs from \(\pi\) down to \(0\) and \(dx=-\sin t\,dt\). Reversing the limits gives
Worked Example: Integrate \(x\arcsin x\) on \([-1,1]\)
Evaluate \(K=\int_{-1}^1x\arcsin x\,dx\). The integrand is real-valued throughout the interval. Choose \(u=\arcsin x\) and \(dv=x\,dx\), giving \(du=dx/\sqrt{1-x^2}\) in the interior and \(v=x^2/2\). Then
The boundary term is
Use \(x=\sin t\), with \(-\pi/2\le t\le\pi/2\), to evaluate the remaining integral:
Hence \(K=\pi/2-\tfrac12(\pi/2)=\pi/4\). As a check, \(x\) and \(\arcsin x\) are both odd, so their product is even. The substitution \(x=\sin t\) gives
in agreement with the integration-by-parts result.
A Common Endpoint Mistake
A common error is to see that \(1/\sqrt{1-x^2}\) becomes unbounded at \(x=1\) and conclude that any integral containing it must diverge. For instance, one might incorrectly write
This treats the value at one endpoint as if it determined the behavior of the entire integral. It does not. The integral is an improper limit, and the singularity is integrable:
The correction is to distinguish the function’s endpoint value from the limit of its integral near that endpoint. The derivative is not finite at \(1\), but the integral of its magnitude over a neighborhood ending at \(1\) is finite.
Practical Checks Before Applying Parts
For a definite integral involving \(\arcsin x\) or \(\arccos x\), first confirm that every \(x\) in the interval lies in \([-1,1]\). Next, note whether either endpoint is \(-1\) or \(1\). If so, use the integration-by-parts formula on truncated intervals and take the limit. Evaluate boundary terms using the finite endpoint values of the inverse-trigonometric function; do not substitute an infinite derivative value at an endpoint.
These checks allow integration by parts to be used safely on intervals that reach the edge of the real domain. The essential point is not to avoid such intervals, but to handle the endpoint limits explicitly.
Check Your Understanding
Use the real domain and endpoint behavior of the inverse-trigonometric functions in your reasoning.
- What is the real domain of \(\arccos x\), and what is \(\arccos(-1)\)?
- Why does the unbounded derivative of \(\arcsin x\) at \(x=1\) not by itself imply that an integral involving that derivative diverges?
- For integration by parts with \(u=\arccos x\), what sign appears in front of the remaining derivative integral, and why?
- Why must an interval for a real integral involving \(\arcsin x\) be contained in \([-1,1]\)?
- Describe how to justify integration by parts when the interval has \(1\) as its right endpoint.
