Choosing and Solving Mixed Radical Integrals

Why the Choice of Substitution Matters

An integral can combine a quadratic radical with factors of \(x\), \(1/x\), or a polynomial expression. The radical often signals a useful trigonometric substitution, but that choice must also simplify the other factors and respect the domain. A substitution that simplifies the square root but leaves a more complicated integrand may not be the best choice.

This tutorial focuses on integrals involving the forms \(\sqrt{a^2-x^2}\) and \(\sqrt{x^2-a^2}\), where \(a>0\). The goal is to select a substitution deliberately, carry every factor through it, return to the original variable with the correct signs, and check the final result by differentiation.

Choosing a Substitution and a Branch

For \(\sqrt{a^2-x^2}\), assume \(-a<x<a\) and take \(\theta=\arcsin(x/a)\), so \(-\pi/2<\theta<\pi/2\). Then \(\cos\theta>0\), and

These identities follow because \(a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2\theta\), and \(\cos\theta\) is positive on this chosen interval. The sign of \(x\), however, is the sign of \(\sin\theta\); it need not be positive.

For a right-triangle picture with positive side lengths, it is simplest to restrict to \(0<x<a\). The triangle then has opposite side \(x\), adjacent side \(\sqrt{a^2-x^2}\), and hypotenuse \(a\). If \(-a<x<0\), the opposite side length is \(|x|\), not \(x\). The signed substitution still gives \(\cot\theta=\sqrt{a^2-x^2}/x\), which is negative in that case.

For \(\sqrt{x^2-a^2}\), the branch \(x>a\) is handled by \(x=a\sec\theta\) with \(0<\theta<\pi/2\). Here \(\tan\theta>0\), so

The radical relation follows from \(\sec^2\theta-1=\tan^2\theta\). Restricting to \(x>a\) ensures the stated positive square root is \(a\tan\theta\). Integrals on other domain intervals require their own branch and sign choices.

After making the substitution, simplify the whole integrand, not just the radical. For example, a factor \(1/x^2\) becomes a reciprocal trigonometric factor under \(x=a\sin\theta\); it must be carried along with both the radical and \(dx\).

Worked Example: A Radical Divided by \(x\)

Worked Example: A Radical Divided by \(x^2\)

Worked Example: Combining Polynomial and Reciprocal Factors

Why the Substitution Works

The substitution is useful because it converts the quadratic expression under the radical into a trigonometric identity and also supplies a differential that often cancels a trigonometric factor. For the sine substitution, the complete change of variables is

On the principal arcsine branch, \(\cos\theta>0\), so taking the positive square root gives \(\sqrt{a^2-x^2}=a\cos\theta\), not an expression with an undetermined sign. Similarly, for the secant substitution on \(x>a\),

Because \(0<\theta<\pi/2\), both \(\sec\theta\) and \(\tan\theta\) are positive, giving \(\sqrt{x^2-a^2}=a\tan\theta\). These branch conditions justify the radical formulas used in the examples.

A Sign and Triangle Pitfall

A right triangle records positive lengths, whereas a trigonometric substitution can involve a negative \(x\). For example, if \(x=-2\) and \(a=4\), then \(\theta=\arcsin(-2/4)=-\pi/6\). The radical is \(\sqrt{16-(-2)^2}=2\sqrt3\).

A wrong triangle description would call the opposite side \(-2\). That cannot be a side length. The correct triangle has opposite side \(|x|=2\), adjacent side \(2\sqrt3\), and hypotenuse \(4\). The angle is negative, which preserves the sign in the trigonometric ratios:

Thus the triangle supplies side lengths, while the signed substitution determines whether a trigonometric ratio is positive or negative. On an interval where \(x>0\), the simpler positive-side triangle used in the first and third examples is valid.

Why Verification Is Part of the Method

Back-substitution can introduce errors even when the trigonometric integration is correct. Common sources include losing a factor of \(a\), replacing a signed ratio with a positive side-length ratio, or simplifying a logarithm incorrectly. Differentiating the final expression tests all of these decisions at once.

The check must use the original variable and the original domain. It is not enough that an expression looks plausible or agrees at one sample value: its derivative must equal the integrand throughout the interval where the formula is claimed. When a domain has multiple intervals, an antiderivative is considered separately on each interval, and the constant of integration may differ from one interval to another.