Another Inverse Function Times a Polynomial
The integrand \(x\arcsin x\) pairs a polynomial with an inverse trigonometric function, just like \(x\arctan x\) in an earlier tutorial. The derivative of \(\arcsin x\) is \(\frac{1}{\sqrt{1-x^2}}\), which is algebraic, so the inverse function is assigned to \(u\). This time the leftover integral contains a square root, and the method for it is trigonometric substitution.
The domain matters. The function \(\arcsin x\) is defined only for \(-1\le x\le1\), and its derivative exists only on \((-1,1)\). All formulas below are therefore stated on the open interval \((-1,1)\), and definite integrals extend to the endpoints because the integrand is continuous on \([-1,1]\).
Deriving the Antiderivative
Let \(u=\arcsin x\) and \(dv=x\,dx\). Then \(du=\frac{dx}{\sqrt{1-x^2}}\) and \(v=\frac{x^2}{2}\), so
For the remaining integral, let \(x=\sin\theta\) with \(-\frac\pi2<\theta<\frac\pi2\). Then \(dx=\cos\theta\,d\theta\) and \(\sqrt{1-x^2}=\cos\theta>0\). The integral becomes
Returning to \(x\): \(\theta=\arcsin x\), \(\sin\theta=x\), and \(\cos\theta=\sqrt{1-x^2}\). So \(\int\frac{x^2}{\sqrt{1-x^2}}\,dx=\frac12\arcsin x-\frac12x\sqrt{1-x^2}\). Multiplying by \(-\frac12\) and combining with the first term,
Checking by Differentiation
Differentiate the first term: \(\frac{2x^2-1}{4}\cdot\frac{1}{\sqrt{1-x^2}}+x\arcsin x\). Differentiate the second term with the product rule: \(\frac14\sqrt{1-x^2}-\frac{x^2}{4\sqrt{1-x^2}}\). Place the three algebraic pieces over \(4\sqrt{1-x^2}\): the numerator is \(2x^2-1+(1-x^2)-x^2=0\). Only \(x\arcsin x\) remains, as required.
Worked Examples
Worked Example: From 0 to 1
Evaluate \(\int_0^1x\arcsin x\,dx\).
At \(x=1\): \(\frac{2-1}{4}\cdot\frac\pi2+0=\frac\pi8\). At \(x=0\) the antiderivative is \(0\). So
The square-root term vanishes at both endpoints, so only the arcsine term contributes. This is typical of integrals over the full interval.
Worked Example: From 0 to 1/2
Evaluate \(\int_0^{1/2}x\arcsin x\,dx\).
At \(x=\frac12\): \(2x^2-1=-\frac12\), \(\arcsin\frac12=\frac\pi6\), and \(x\sqrt{1-x^2}=\frac12\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{4}\). Thus the value is
The answer is small and positive, as it should be: the integrand is non-negative and at most \(\frac12\cdot\frac\pi6\approx0.26\) on a short interval, and it starts at zero.
Worked Example: Symmetric Limits
Evaluate \(\int_{-1}^{1}x\arcsin x\,dx\).
The product of two odd functions is even, so the integral is twice the integral over \([0,1]\):
As with \(x\arctan x\), the integrand does not cancel on a symmetric interval, because it is even, not odd.
Worked Example: Upper Limit √3/2
Evaluate \(\int_0^{\sqrt3/2}x\arcsin x\,dx\).
At \(x=\frac{\sqrt3}{2}\): \(2x^2-1=\frac12\), \(\arcsin x=\frac\pi3\), and \(x\sqrt{1-x^2}=\frac{\sqrt3}{2}\cdot\frac12=\frac{\sqrt3}{4}\). Therefore
Both pieces contribute here, showing that the algebraic term cannot be ignored away from the endpoints \(0\) and \(1\).
A Reliable Procedure
Take \(u=\arcsin x\) and \(dv=x\,dx\).
Write \(\frac{x^2}{2}\arcsin x-\frac12\int\frac{x^2}{\sqrt{1-x^2}}\,dx\).
Let \(x=\sin\theta\), so the integral becomes \(\int\sin^2\theta\,d\theta\).
Use \(\theta=\arcsin x\), \(\sin\theta=x\), \(\cos\theta=\sqrt{1-x^2}\).
Merge the two arcsine terms and differentiate to verify.
Common Mistakes
The first common error is a sign slip. The leftover integral is multiplied by \(-\frac12\), so the arcsine contribution is \(-\frac14\arcsin x\) and the square-root term is \(+\frac14x\sqrt{1-x^2}\). Dropping the minus sign reverses both. A second error is leaving the answer in terms of \(\theta\) or forgetting that \(\cos\theta=\sqrt{1-x^2}\) requires \(\cos\theta>0\), which holds for \(-\frac\pi2<\theta<\frac\pi2\).
A third error is to treat the integral on a symmetric interval as zero. The integrand \(x\arcsin x\) is even, so the integral over \([-a,a]\) is twice the integral over \([0,a]\), not zero.
A Cleverer Choice of v
As with the arctangent version, choose \(v=\frac{x^2-1}{2}\) instead of \(\frac{x^2}{2}\). Then the leftover integral is \(-\int\frac{x^2-1}{2\sqrt{1-x^2}}\,dx=\frac12\int\sqrt{1-x^2}\,dx\), which is the area of a circular segment: \(\frac12\cdot\frac12\bigl(x\sqrt{1-x^2}+\arcsin x\bigr)\). The result is \(\frac{x^2-1}{2}\arcsin x+\frac14x\sqrt{1-x^2}+\frac14\arcsin x\), and collecting arcsine terms again gives \(\frac{2x^2-1}{4}\arcsin x+\frac{x\sqrt{1-x^2}}{4}\). Both routes agree, and this one avoids a separate trigonometric substitution.
Quick Reference
| Integrand | Antiderivative |
|---|---|
| \(\arcsin x\) | \(x\arcsin x+\sqrt{1-x^2}\) |
| \(x\arcsin x\) | \(\frac{2x^2-1}{4}\arcsin x+\frac{x\sqrt{1-x^2}}{4}\) |
| \(\frac{x^2}{\sqrt{1-x^2}}\) | \(\frac12\arcsin x-\frac12x\sqrt{1-x^2}\) |
| \(\sqrt{1-x^2}\) | \(\frac12\bigl(x\sqrt{1-x^2}+\arcsin x\bigr)\) |
Practice Problems with Answers
Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.
- Evaluate \(\int_0^1x\arcsin x\,dx\). Hint: the square-root term vanishes at \(0\) and \(1\). Answer: \(\frac\pi8\)
- Find an antiderivative of \(4x\arcsin x\). Hint: multiply the result by \(4\). Answer: \((2x^2-1)\arcsin x+x\sqrt{1-x^2}\)
- Evaluate \(\int_0^{1/2}x\arcsin x\,dx\). Hint: evaluate at \(x=\frac12\), where \(\arcsin x=\frac\pi6\). Answer: \(\frac{\sqrt3}{16}-\frac\pi{48}\)
- Evaluate \(\int_{-1}^1x\arcsin x\,dx\). Hint: double the integral over \([0,1]\). Answer: \(\frac\pi4\)
Interpreting the Result
The antiderivative can be read geometrically. The term \(\frac{x\sqrt{1-x^2}}{4}\) is half the area of the right triangle with legs \(x\) and \(\sqrt{1-x^2}\) inscribed in the unit circle, and the arcsine terms measure sector angles. This is no accident: the integral \(\int\sqrt{1-x^2}\,dx\), the area under a quarter circle, hides inside the problem, which is why a circular-segment formula appears in the cleverer derivation above.
The behavior at the endpoints is also informative. At \(x=1\) the square-root term disappears and the arcsine takes its maximum value \(\frac\pi2\). The integrand \(x\arcsin x\) then equals \(\frac\pi2\), and its integral from \(0\) to \(1\) is \(\frac\pi8\approx0.393\), exactly one quarter of the rectangle \([0,1]\times[0,\frac\pi2]\) of area \(\frac\pi2\). The integrand is small near zero and rises steeply near \(x=1\), where the arcsine has a vertical tangent, so the area below the curve is well below half the rectangle, as the value \(0.25\) of the ratio indicates.
A Numerical Spot Check
Estimate \(\int_0^1x\arcsin x\,dx\) with the midpoint rule on two subintervals. The midpoints are \(0.25\) and \(0.75\), where \(x\arcsin x\) takes the values \(0.25\cdot0.2527=0.0632\) and \(0.75\cdot0.8481=0.6361\). The estimate is \(0.5\cdot(0.0632+0.6361)=0.350\), compared with the exact value \(0.393\). A coarse rule underestimates here because the curve is steep and convex near \(x=1\), so the agreement within about ten percent is exactly what one expects.
Why the Sine Substitution Is the Natural Choice
The radical \(\sqrt{1-x^2}\) has the form \(\sqrt{a^2-x^2}\) with \(a=1\). The Pythagorean identity \(1-\sin^2\theta=\cos^2\theta\) turns the radical into a plain trigonometric function, and the differential \(dx=\cos\theta\,d\theta\) supplies a factor that cancels the radical in the denominator. After the cancellation only \(\int\sin^2\theta\,d\theta\) remains, which a half-angle identity handles at once. A direct attempt to integrate \(\frac{x^2}{\sqrt{1-x^2}}\) by algebra alone has no comparable shortcut.
This pairing of an inverse sine in the parts step with a sine substitution in the second step is common. The inverse function introduced the radical, and the substitution that undoes the inverse function, \(x=\sin\theta\), removes it again. The same pairing appears with inverse secant and tangent in later tutorials.
Where This Fits in the Course
This tutorial belongs to the module “Integration by Parts I”. The tutorial just before it is “Integrating x Squared Times The Inverse Tangent,” and the one that follows is “Definite Integrals Involving The Inverse Tangent.” It combines integration by parts with the trigonometric substitution you will study in depth later, so it is a good preview of how the two techniques cooperate in a single problem.
Check Your Understanding
Use the derivation and examples above to answer these questions.
- Why must the answer be restricted to \(-1<x<1\) as an indefinite integral?
- What does \(\sqrt{1-x^2}\) equal when \(x=\sin\theta\) and \(|\theta|<\frac\pi2\)?
- Evaluate \(\int_0^1x\arcsin x\,dx\).
- Is \(x\arcsin x\) even, odd, or neither?
- Differentiate the final antiderivative and confirm that the algebraic terms cancel.
