Integrating the Square Root of a^2 + x^2

Why This Radical Calls for a Tangent Substitution

The expression \(\sqrt{a^2+x^2}\) is positive for every real \(x\) when \(a>0\). Its sum of squares suggests the identity \(1+\tan^2\theta=\sec^2\theta\). Setting \(x=a\tan\theta\) turns the radical into a multiple of \(\sec\theta\), and the differential contributes another factor of \(\sec^2\theta\).

The resulting trigonometric integral involves \(\sec^3\theta\). Integration by parts reduces that integral to a logarithm and an algebraic term. A right triangle then expresses both terms using \(x\) and \(\sqrt{a^2+x^2}\).

Set Up the Tangent Substitution

For \(a>0\), choose

This interval is useful because \(\sec\theta>0\) throughout it. Differentiating the substitution gives \(dx=a\sec^2\theta\,d\theta\), while the identity \(1+\tan^2\theta=\sec^2\theta\) gives

The positive square root is \(a\sec\theta\), rather than \(-a\sec\theta\), because \(a>0\) and \(\sec\theta>0\) on the chosen interval. Therefore, if \(I\) denotes the integral in the new variable,

Both factors of \(a\) matter: one comes from the radical and the other from \(dx\).

Derive the Secant-Cubed Integral

To evaluate \(\int\sec^3\theta\,d\theta\), write the integrand as \(\sec\theta\sec^2\theta\) and use integration by parts. Choose \(u=\sec\theta\) and \(dv=\sec^2\theta\,d\theta\). Then \(du=\sec\theta\tan\theta\,d\theta\) and \(v=\tan\theta\). If \(J=\int\sec^3\theta\,d\theta\), integration by parts yields

Now use \(\tan^2\theta=\sec^2\theta-1\). The remaining integral becomes \(J-\int\sec\theta\,d\theta\), so

Rearranging gives \(2J=\sec\theta\tan\theta+\int\sec\theta\,d\theta\). On the chosen interval, \(\sec\theta+\tan\theta>0\): indeed, \(\sec\theta>|\tan\theta|\) follows from \(\sec^2\theta-\tan^2\theta=1\). Thus the standard logarithmic antiderivative of \(\sec\theta\) can be written without absolute-value bars here. Differentiating \(\ln(\sec\theta+\tan\theta)\) gives \(\sec\theta\), so

Return to \(x\) with a Right Triangle

From \(x=a\tan\theta\), we have \(\tan\theta=x/a\). To find \(\sec\theta\), use a right triangle with adjacent side \(a\), opposite side \(|x|\), and hypotenuse \(\sqrt{a^2+x^2}\). The absolute value labels a side length; the sign of \(x\) is retained in \(\tan\theta=x/a\).

The triangle gives \(\sec\theta=\sqrt{a^2+x^2}/a\). Consequently,

The numerator \(\sqrt{a^2+x^2}+x\) is positive: \(\sqrt{a^2+x^2}>|x|\) when \(a>0\). Substituting these expressions into \(a^2J\) gives the antiderivative.

The ratio inside the logarithm is positive. It is also dimensionless when \(a\) and \(x\) have the same units, as required for the logarithm of a physical quantity.

Worked Examples

A Reliable Derivative Check

The examples illustrate why it is useful to verify an antiderivative independently. For general \(a>0\), let \(R=\sqrt{a^2+x^2}\) and differentiate the proposed formula. Since \(R'=x/R\) and

we obtain

This confirms the antiderivative. It also confirms that the logarithm coefficient must be \(a^2\), not \(a\), and that the factor \(1/2\) applies to both terms.

Common Pitfalls

Dropping a factor of \(a\). The radical contributes \(a\sec\theta\), and the differential contributes \(a\sec^2\theta\,d\theta\). Their product is \(a^2\sec^3\theta\,d\theta\), not \(a\sec^3\theta\,d\theta\). For \(a=4\), the multiplier is \(16\), as the second example shows.

Using \(|x|\) instead of \(x\) in the tangent ratio. A triangle side is a nonnegative length, so its opposite side can be labelled \(|x|\). But the substitution says \(\tan\theta=x/a\), including when \(x<0\). Replacing that signed ratio with \(|x|/a\) loses the sign needed to recover \(x\).

Taking the wrong square-root branch. The identity gives \(\sqrt{a^2\sec^2\theta}=a|\sec\theta|\) in general. The chosen range \(-\pi/2<\theta<\pi/2\) ensures \(\sec\theta>0\), so the radical is \(a\sec\theta\). The range is part of the justification, not just a convenience.

When This Result Is Useful

The formula is useful whenever an integral contains precisely the radical \(\sqrt{a^2+x^2}\) as its integrand. Its structure records the two main contributions: an algebraic term \(x\sqrt{a^2+x^2}\) and a logarithmic term. Keeping the parameter \(a\) visible helps track the scaling and prevents losing a factor of \(a^2\).

For a quick check, differentiate the result rather than relying only on the substitution steps. In particular, the derivative of the logarithm is \(1/\sqrt{a^2+x^2}\), and the algebraic term supplies the remaining part needed to recover the radical. The logarithm's argument is positive for all real \(x\) when \(a>0\), so no absolute value is needed in this formula.