Why a Tangent Substitution Fits This Radical
The expression \(a^2+x^2\), with \(a>0\), has the form of a sum of two squares. A right triangle relates that sum to the square of its hypotenuse. The tangent function relates the opposite and adjacent sides, while the secant function relates the hypotenuse and adjacent side. These relationships suggest a substitution that simplifies the radical and gives a precise change of variable for real-variable integrals.
The substitution is useful only when its signs and angle range are handled correctly. In particular, a square root is nonnegative, so the angle must be chosen where the secant is positive.
Deriving the Substitution from a Right Triangle
Suppose first that \(x\geq 0\). Draw a right triangle whose adjacent leg has length \(a\), whose opposite leg has length \(x\), and whose acute angle is \(\theta\). By the Pythagorean theorem, the hypotenuse has length \(\sqrt{a^2+x^2}\). The definitions of tangent and secant then give
Rearranging both relations produces the desired formulas. For negative \(x\), use a triangle with opposite side length \(|x|\) and take \(\theta\) to be negative. The tangent relation then remains \(\tan\theta=x/a\), while the hypotenuse remains positive.
These identities require an angle range that makes the tangent represent every real \(x\) and keeps the secant positive. We choose
On this interval, tangent is strictly increasing and takes every real value exactly once. Also, \(\cos\theta>0\), so \(\sec\theta>0\). Thus the second identity agrees with the nonnegative square root, including when \(x\) is negative.
Complete Justification of the Radical Identity
The triangle motivates the formulas, and a direct calculation proves the radical identity for every real \(\theta\) in the chosen interval. Starting with \(x=a\tan\theta\), square the expression inside the radical and use the trigonometric identity \(1+\tan^2\theta=\sec^2\theta\):
The absolute value in the last line is necessary because \(\sqrt{u^2}=|u|\). On the chosen interval, \(\cos\theta>0\), so \(\sec\theta>0\) and \(|\sec\theta|=\sec\theta\). Therefore
This argument also explains why simply dropping the absolute value without checking the angle would be unsafe. The interval is part of the substitution, not an optional convention.
The Differential and the Inverse Change of Variable
Differentiate \(x=a\tan\theta\) with respect to \(\theta\). Since \(a\) is constant and \(\frac{d}{d\theta}\tan\theta=\sec^2\theta\), the differential is
This factor is positive throughout the chosen interval, because \(a>0\) and \(\sec^2\theta>0\). Hence increasing \(\theta\) corresponds to increasing \(x\), which preserves the order of endpoints when changing limits in a definite integral.
The inverse relation is \(\theta=\arctan(x/a)\), where \(\arctan\) takes values in \((-\pi/2,\pi/2)\). Differentiating it gives a useful check:
From \(\tan\theta=x/a\), we also have \(\sec^2\theta=1+x^2/a^2=(a^2+x^2)/a^2\). Therefore \(dx/d\theta=a\sec^2\theta=(a^2+x^2)/a\), the reciprocal of \(d\theta/dx\), as required.
Applying the Substitution Consistently
When an integral contains \(\sqrt{a^2+x^2}\), replacing the radical alone is not enough. Every occurrence of \(x\) must be rewritten using \(a\tan\theta\), the radical using \(a\sec\theta\), and the differential using \(a\sec^2\theta\,d\theta\). For a definite integral, the endpoints must also be translated using \(\theta=\arctan(x/a)\).
Set \(\theta=\arctan(x/a)\), so \(-\pi/2<\theta<\pi/2\).
Use \(x=a\tan\theta\) and \(\sqrt{a^2+x^2}=a\sec\theta\).
Use \(dx=a\sec^2\theta\,d\theta\).
For each endpoint \(x=c\), use \(\theta=\arctan(c/a)\), keeping the angle in the stated interval.
Worked Examples
Worked Example: A Negative Value of \(x\)
Let \(a=8\) and \(x=-6\). Find the corresponding angle relation, evaluate the radical using the substitution, and check the differential factor at that angle.
First, \(\tan\theta=x/a=-6/8=-3/4\). The angle is \(\theta=\arctan(-3/4)\), which lies in \((-\pi/2,\pi/2)\). A triangle with adjacent side \(8\) and opposite side length \(6\) has hypotenuse
Because the angle is negative, \(\tan\theta=-3/4\), while \(\sec\theta\) is positive. The triangle gives \(\sec\theta=10/8=5/4\). Thus the substitution predicts
The differential factor at this angle is
As a check, the inverse derivative is \(d\theta/dx=8/(64+36)=8/100=2/25\). Its reciprocal is \(25/2\), matching \(dx/d\theta\). The radical and derivative are therefore consistent.
Worked Example: Transforming a Differential Expression
Take \(a=5\) and \(x=12\). Express \(\sqrt{a^2+x^2}\,dx\) in terms of \(\theta\), then check both factors.
Here \(\tan\theta=12/5\), so \(\theta=\arctan(12/5)\) lies in the required interval. The triangle has hypotenuse
Thus \(\sec\theta=13/5\), and the radical becomes \(5\sec\theta=5(13/5)=13\). Also,
Multiplying the transformed factors gives
To check, \(d\theta/dx=5/(25+144)=5/169\), so \(dx/d\theta=169/5\), as used. The radical check is \(5\sec\theta=5(13/5)=13=\sqrt{169}\).
Worked Example: Converting Definite-Integral Limits
For \(a=3\), convert the limits \(x=-3\) and \(x=3\) to angle limits. Then state the corresponding general substitution for an integrand \(H(x,\sqrt{9+x^2})\), where \(H\) is any function for which the integral exists.
At the lower endpoint, \(\tan\theta=-3/3=-1\), so \(\theta=-\pi/4\). At the upper endpoint, \(\tan\theta=3/3=1\), so \(\theta=\pi/4\). The transformed variable increases from \(-\pi/4\) to \(\pi/4\), in agreement with \(dx/d\theta=3\sec^2\theta>0\).
Substitute \(x=3\tan\theta\), \(\sqrt{9+x^2}=3\sec\theta\), and \(dx=3\sec^2\theta\,d\theta\). The definite integral becomes
At \(\theta=-\pi/4\), \(x=3(-1)=-3\); at \(\theta=\pi/4\), \(x=3(1)=3\). These checks confirm both endpoint conversions. Also \(\sec(\pm\pi/4)=\sqrt{2}>0\), consistent with the positive square root.
A Common Mistake: Ignoring the Angle Range
A frequent error is to write \(\sqrt{a^2+a^2\tan^2\theta}=a\sec\theta\) for any angle. The calculation actually gives \(a|\sec\theta|\). For example, let \(a=2\) and \(\theta=2\pi/3\). Then \(\tan\theta=-\sqrt{3}\), so \(x=2\tan\theta=-2\sqrt{3}\). The radical is
But \(\sec(2\pi/3)=-2\), so \(a\sec\theta=2(-2)=-4\), which cannot equal a nonnegative square root. The correction is to use the principal angle \(\theta=\arctan(x/a)=-\pi/3\), which is in \((-\pi/2,\pi/2)\). Then \(\sec\theta=2\), and \(a\sec\theta=4\), as required.
Another error is to omit the factor \(a\) when differentiating. If \(x=3\tan\theta\), for instance, then \(dx/d\theta=3\sec^2\theta\), not merely \(\sec^2\theta\). The constant multiplier in the substitution must remain in the differential.
Why the Substitution Matters
The substitution turns the quadratic radical into a trigonometric expression whose sign is controlled by the angle interval. Its derivative supplies the exact factor needed to change variables, and the one-to-one correspondence between real \(x\) and angles in \((-\pi/2,\pi/2)\) makes endpoint conversion unambiguous. These features make it a reliable tool for integrals involving \(a^2+x^2\), provided the radical, differential, and limits are all transformed together.
Check Your Understanding
Use the angle interval and sign conditions carefully in each question.
- For \(a=7\) and \(x=-24\), find \(\tan\theta\), the hypotenuse length, and \(\sec\theta\).
- Why does \(\sqrt{a^2+a^2\tan^2\theta}\) initially simplify to \(a|\sec\theta|\), rather than automatically to \(a\sec\theta\)?
- Find \(dx/d\theta\) when \(x=6\tan\theta\), and state why this derivative is positive on the chosen angle interval.
- Under the substitution with \(a=4\), convert the endpoints \(x=0\) and \(x=4\) to angle limits.
- Identify the error in replacing \(dx\) by \(\sec^2\theta\,d\theta\) when \(x=5\tan\theta\).