Polynomial Weights Over √(a² + x²)
The radical \(\sqrt{a^2+x^2}\) is defined for every real \(x\), which makes it friendlier than \(\sqrt{x^2-a^2}\). Integrals of the form \(\int\frac{x^n}{\sqrt{a^2+x^2}}\,dx\) are solved by the tangent substitution \(x=a\tan\theta\). The identity \(1+\tan^2\theta=\sec^2\theta\) turns the radical into \(a\sec\theta\), and the weights \(x^n\) become powers of \(\tan\theta\). For odd \(n\) the algebraic substitution \(u=\sqrt{a^2+x^2}\) is a simpler alternative.
Deriving the Formulas
Let \(x=a\tan\theta\) with \(-\frac\pi2<\theta<\frac\pi2\). Then \(dx=a\sec^2\theta\,d\theta\) and \(\sqrt{a^2+x^2}=a\sec\theta\), so \(\frac{dx}{\sqrt{a^2+x^2}}=\sec\theta\,d\theta\) and
For \(n=1\), \(\int\tan\theta\sec\theta\,d\theta=\sec\theta\), giving \(a\sec\theta=\sqrt{a^2+x^2}\). For \(n=2\), write \(\tan^2\theta=\sec^2\theta-1\): \(\int(\sec^3\theta-\sec\theta)\,d\theta=\frac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)-\ln|\sec\theta+\tan\theta|=\frac12(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|)\). With \(\sec\theta=\frac{\sqrt{a^2+x^2}}{a}\) and \(\tan\theta=\frac xa\), this yields the second formula. For \(n=3\), write \(\tan^3\theta\sec\theta=(\sec^2\theta-1)\sec\theta\tan\theta\), whose integral is \(\frac{\sec^3\theta}{3}-\sec\theta\); multiplying by \(a^3\) gives the third formula.
The Algebraic Shortcut
For odd \(n\), let \(u=\sqrt{a^2+x^2}\). Then \(u\,du=x\,dx\) and \(x^2=u^2-a^2\). For \(n=3\), the integral becomes \(\int\frac{(u^2-a^2)u\,du}{u}=\int(u^2-a^2)\,du=\frac{u^3}{3}-a^2u\), which matches the trigonometric answer. The shortcut works because the odd power of \(x\) provides the factor \(x\,dx\).
A Reduction Relation
Differentiating \(x^{n-1}\sqrt{a^2+x^2}\) gives \(\frac{n\,x^n+(n-1)a^2x^{n-2}}{\sqrt{a^2+x^2}}\), and therefore
This relation reduces any power to \(n=0\) or \(n=1\). It also serves as an independent check on the formulas above: \(n=2\) gives \(\frac{x\sqrt{a^2+x^2}}{2}-\frac{a^2}{2}\int\frac{dx}{\sqrt{a^2+x^2}}\), and the last integral is \(\ln(x+\sqrt{a^2+x^2})\).
Worked Examples
Worked Example: First Power
Evaluate \(\int_0^{\sqrt3}\frac{x\,dx}{\sqrt{1+x^2}}\).
The antiderivative is \(\sqrt{1+x^2}\). At \(x=\sqrt3\) it equals \(2\), and at \(x=0\) it equals \(1\). Therefore
The integrand rises from \(0\) to \(\frac{\sqrt3}{2}\approx0.866\), so an average near \(0.58\) over an interval of length \(1.73\) gives an area of about \(1\).
Worked Example: Second Power
Evaluate \(\int_0^1\frac{x^2\,dx}{\sqrt{1+x^2}}\).
With \(a=1\), the antiderivative is \(\frac x2\sqrt{1+x^2}-\frac12\ln(x+\sqrt{1+x^2})\). At \(x=1\): \(\frac{\sqrt2}{2}-\frac12\ln(1+\sqrt2)\). At \(x=0\): \(0-\frac12\ln1=0\). Hence
The value is small because the integrand \(\frac{x^2}{\sqrt{1+x^2}}\) is at most \(\frac{1}{\sqrt2}\) on \([0,1]\) and near zero for small \(x\).
Worked Example: Third Power
Evaluate \(\int_0^{\sqrt3}\frac{x^3\,dx}{\sqrt{1+x^2}}\).
With \(u=\sqrt{1+x^2}\), the limits change from \(u=1\) to \(u=2\), and the integral becomes \(\int_1^2(u^2-1)\,du=\bigl[\frac{u^3}{3}-u\bigr]_1^2=\bigl(\frac83-2\bigr)-\bigl(\frac13-1\bigr)\). Therefore
Changing the limits with the substitution is cleaner than back-substituting, since no radical appears in the final arithmetic.
Worked Example: Radius Two
Evaluate \(\int_0^2\frac{x^2\,dx}{\sqrt{4+x^2}}\).
With \(a=2\), the antiderivative is \(\frac x2\sqrt{4+x^2}-2\ln(x+\sqrt{4+x^2})\). At \(x=2\): \(\sqrt8-2\ln(2+\sqrt8)=2\sqrt2-2\ln(2+2\sqrt2)\). At \(x=0\): \(-2\ln2\). Subtracting and combining logarithms,
The combination \(\ln(2+2\sqrt2)-\ln2=\ln(1+\sqrt2)\) is the key simplification.
Worked Example: A Fourth Power from the Reduction Relation
Evaluate \(\int_0^1\frac{x^4\,dx}{\sqrt{1+x^2}}\).
With \(n=4\) and \(a=1\), the relation gives \(\frac{x^3\sqrt{1+x^2}}{4}-\frac34\int\frac{x^2\,dx}{\sqrt{1+x^2}}\). At \(x=1\) the first term is \(\frac{\sqrt2}{4}\), and the second integral over \([0,1]\) is \(\frac{\sqrt2}{2}-\frac12\ln(1+\sqrt2)\). Thus
The answer is positive but small, as expected for an integrand that behaves like \(x^4\) near zero.
A Reliable Procedure
For odd \(n\), try \(u=\sqrt{a^2+x^2}\).
Set \(x=a\tan\theta\), so the integrand is \(a^n\tan^n\theta\sec\theta\,d\theta\).
Use \(\tan^2\theta=\sec^2\theta-1\) to reach powers of secant.
Use \(\sec\theta=\frac{\sqrt{a^2+x^2}}{a}\) and \(\tan\theta=\frac xa\).
Differentiate, or compare with the reduction relation.
Common Mistakes
The sign in the second and third formulas is the opposite of the corresponding formulas for \(\sqrt{x^2-a^2}\). Writing \(+\frac{a^2}{2}\ln\) in the second formula is the most common error. Another is to forget that \(\sqrt{a^2+x^2}\) does not simplify to \(a+x\). A third error is to apply the shortcut \(u=\sqrt{a^2+x^2}\) to even powers. Finally, when evaluating a definite integral, remember that \(\ln(x+\sqrt{a^2+x^2})\) at \(x=0\) equals \(\ln a\), not zero, unless \(a=1\); forgetting this changes the answer.
Quick Reference
| n | Antiderivative of \(\frac{x^n}{\sqrt{a^2+x^2}}\) |
|---|---|
| 1 | \(\sqrt{a^2+x^2}\) |
| 2 | \(\frac x2\sqrt{a^2+x^2}-\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})\) |
| 3 | \(\frac{(a^2+x^2)^{3/2}}{3}-a^2\sqrt{a^2+x^2}\) |
Interpreting the Results
For large \(x\), \(\frac{x^n}{\sqrt{a^2+x^2}}\approx x^{n-1}\), so the antiderivatives grow like \(\frac{x^n}{n}\). The formulas agree: the second is about \(\frac{x^2}{2}\), since the logarithm is lower order, and the third is about \(\frac{x^3}{3}\). Unlike the case \(\sqrt{x^2-a^2}\), there is no singularity for any real \(x\), so every definite integral is proper.
Symmetry on Negative Intervals
The integrand \(\frac{x^n}{\sqrt{a^2+x^2}}\) is odd for odd \(n\) and even for even \(n\). Over a symmetric interval \([-b,b]\), odd cases give zero, while even cases give twice the integral over \([0,b]\). For example, \(\int_{-1}^{1}\frac{x^2\,dx}{\sqrt{1+x^2}}=\sqrt2-\ln(1+\sqrt2)\approx0.533\), and \(\int_{-1}^{1}\frac{x^3\,dx}{\sqrt{1+x^2}}=0\).
Practice Problems with Answers
Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.
- Evaluate \(\int_0^{\sqrt3}\frac{x\,dx}{\sqrt{1+x^2}}\). Hint: the antiderivative is \(\sqrt{1+x^2}\). Answer: \(1\)
- Evaluate \(\int_0^1\frac{x^2\,dx}{\sqrt{1+x^2}}\). Hint: use the second formula with \(a=1\). Answer: \(\frac{\sqrt2}{2}-\frac12\ln(1+\sqrt2)\)
- Evaluate \(\int_0^{\sqrt3}\frac{x^3\,dx}{\sqrt{1+x^2}}\). Hint: let \(u=\sqrt{1+x^2}\). Answer: \(\frac43\)
- Evaluate \(\int_0^2\frac{x^2\,dx}{\sqrt{4+x^2}}\). Hint: combine the logarithms. Answer: \(2\sqrt2-2\ln(1+\sqrt2)\)
A Numerical Spot Check
Check \(\int_0^1\frac{x^2\,dx}{\sqrt{1+x^2}}\approx0.266\) with Simpson's rule on three points. The heights are \(0\), \(\frac{0.25}{\sqrt{1.25}}\approx0.2236\), and \(\frac{1}{\sqrt2}\approx0.7071\). Then \(\frac{0.5}{3}(0+4\cdot0.2236+0.7071)\approx0.2669\), which agrees with the exact value \(0.2664\) to three digits. A comparison like this takes a minute and catches a wrong sign on the logarithm, since flipping it would change the answer to \(0.7071+0.4407=1.15\).
Why the Sign of the Logarithm Differs
For \(\sqrt{x^2-a^2}\), the second formula had a plus sign, and here it has a minus sign. The reason is visible in the trigonometric integrals: \(\tan^2\theta\sec\theta=\sec^3\theta-\sec\theta\) in both cases, but the reduction of \(\sec^3\theta\) combines with \(-\sec\theta\) to give \(\frac12(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|)\). In the secant substitution of the previous tutorial, the weight was \(\sec^2\theta\) times \(\sec\theta\), which has no \(-\sec\theta\) to subtract. So the sign pattern follows from which integrand \(\tan^n\theta\sec\theta\) or \(\sec^{n+1}\theta\) appears.
Geometric Meaning
The curve \(y=\sqrt{a^2+x^2}\) is the upper branch of the hyperbola \(y^2-x^2=a^2\). The integral \(\int_0^x\frac{t^2\,dt}{\sqrt{a^2+t^2}}\) can be interpreted as the difference between a triangle area and a hyperbolic sector, which explains the minus sign: a sector of angle \(\operatorname{arcsinh}\frac xa\) is subtracted from the triangle \(\frac x2\sqrt{a^2+x^2}\). In hyperbolic notation, \(\ln(x+\sqrt{a^2+x^2})=\operatorname{arcsinh}\frac xa+\ln a\), so the second formula reads \(\frac x2\sqrt{a^2+x^2}-\frac{a^2}{2}\operatorname{arcsinh}\frac xa\) up to a constant.
Why No Domain Restriction Appears
The substitution \(x=a\tan\theta\) maps the interval \((-\frac\pi2,\frac\pi2)\) onto the entire real line, so every real \(x\) is covered. In contrast, \(x=a\sin\theta\) or \(x=a\sec\theta\) cover only part of the line. This is why the results hold for all real \(x\), and why the logarithm \(\ln(x+\sqrt{a^2+x^2})\) needs no absolute value: since \(\sqrt{a^2+x^2}>|x|\), the argument is always positive.
Where This Fits in the Course
This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Integrating the Reciprocal Square Root of a^2 + x^2,” and the one that follows is “Integrals With x√(a^2 + x^2) in the Denominator.” It parallels the secant case for \(\sqrt{x^2-a^2}\), with the sign of the logarithm term reversed. It also prepares the integrals with an extra factor of \(\frac1x\) that come next, where the logarithm appears in a different guise. Keep the reduction relation in mind, since it gives a second route to every result and lets you check your work independently.
Check Your Understanding
Use the derivation and examples above to answer these questions.
- Why does \(x=a\tan\theta\) turn \(\sqrt{a^2+x^2}\) into \(a\sec\theta\)?
- For which values of \(n\) does \(u=\sqrt{a^2+x^2}\) work directly?
- Evaluate \(\int_0^{\sqrt3}\frac{x\,dx}{\sqrt{1+x^2}}\).
- Why is \(\int_{-1}^{1}\frac{x^3\,dx}{\sqrt{1+x^2}}=0\)?
- Differentiate \(\sqrt{a^2+x^2}\) and confirm the first formula.
