Why This Integral Needs Care
An integral with \(x\sqrt{a^2-x^2}\) in the denominator has two features that matter at once: the square root suggests a trigonometric substitution, while the factor \(x\) remains in the transformed denominator. The substitution must therefore be carried through with its scale factors intact.
Throughout, let \(a>0\) be a constant. The real-valued integrand \(1/[x\sqrt{a^2-x^2}]\) is defined only when \(-a<x<a\) and \(x\ne0\). Thus its domain consists of two separate intervals, \((-a,0)\) and \((0,a)\). We will find an antiderivative on either interval.
Set Up the Sine Substitution
For the radical \(a^2-x^2\), take \(x=a\sin\theta\). Choose \(-\pi/2<\theta<\pi/2\), so \(\cos\theta>0\). This angle range ensures that taking the square root does not introduce an ambiguous sign:
The last equality follows because \(a>0\) and \(\cos\theta>0\). Substituting all three expressions into the integrand gives
Both factors of \(a\) in the denominator matter. After cancelling \(\cos\theta\), one factor of \(a\) remains below the integral. Losing that factor would make the eventual antiderivative incorrect.
Integrate and Return to \(x\)
To integrate \(\csc\theta\), one can verify that \(\ln|\csc\theta-\cot\theta|\) is an antiderivative. Indeed, using \((\csc\theta)'=-\csc\theta\cot\theta\) and \((\cot\theta)'=-\csc^2\theta\),
In the last equality, the numerator factors as \(\csc\theta(\csc\theta-\cot\theta)\). The logarithm is defined here because \(\theta\ne0\), corresponding to \(x\ne0\).
Now use a right triangle to rewrite the trigonometric expression. The hypotenuse is \(a\), the signed opposite side corresponds to \(x\), and the adjacent side is \(\sqrt{a^2-x^2}\). Equivalently, the sine and cosine relations give
These relations keep the sign of \(x\) when \(\theta<0\); the adjacent side, and therefore the square root, remains positive. Consequently,
For the second equality, multiply numerator and denominator of \((a-\sqrt{a^2-x^2})/x\) by \(a+\sqrt{a^2-x^2}\). The numerator becomes \(a^2-(a^2-x^2)=x^2\), and cancelling \(x\) is permitted because \(x\ne0\). Combining the substitution, trigonometric antiderivative, and triangle relation yields the result.
Complete Derivation
Assume \(a>0\), restrict to \(-a<x<a\) with \(x\ne0\), and set \(x=a\sin\theta\) for \(-\pi/2<\theta<\pi/2\).
Then \(dx=a\cos\theta\,d\theta\) and \(\sqrt{a^2-x^2}=a\cos\theta\), since \(\cos\theta>0\).
Substitution and cancellation give \((1/a)\int\csc\theta\,d\theta\).
Since the derivative of \(\ln|\csc\theta-\cot\theta|\) is \(\csc\theta\), and \(\csc\theta-\cot\theta=x/(a+\sqrt{a^2-x^2})\), the stated logarithm follows.
This derivation also shows why the method works: the sine substitution turns the square root into a cosine factor that cancels the cosine in \(dx\), leaving the cosecant integral. The triangle relations then express its antiderivative entirely in the original variable.
Worked Examples
Worked Example: A Radical with \(a=5\)
Evaluate \(\displaystyle\int\frac{dx}{x\sqrt{25-x^2}}\) on either interval \((-5,0)\) or \((0,5)\).
Set \(x=5\sin\theta\), with \(-\pi/2<\theta<\pi/2\). Then \(dx=5\cos\theta\,d\theta\) and \(\sqrt{25-x^2}=5\cos\theta\). Therefore,
The triangle relations give \(\csc\theta=5/x\) and \(\cot\theta=\sqrt{25-x^2}/x\). Hence \(\csc\theta-\cot\theta=x/(5+\sqrt{25-x^2})\), and
To check, write \(s=\sqrt{25-x^2}\), so \(s'=-x/s\). Then
In the final equality, \(s(5+s)+x^2=5s+s^2+x^2=5s+25=5(5+s)\), so the derivative is exactly \(1/[x\sqrt{25-x^2}]\).
Worked Example: A Numerator That Matches \(a\)
Evaluate \(\displaystyle\int\frac{7\,dx}{x\sqrt{49-x^2}}\) for \(0<|x|<7\). Use \(x=7\sin\theta\), \(dx=7\cos\theta\,d\theta\), and \(\sqrt{49-x^2}=7\cos\theta\):
Here the original numerator \(7\) cancels the factor \(1/7\) that would otherwise remain. Integrating and using \(\csc\theta=7/x\) and \(\cot\theta=\sqrt{49-x^2}/x\) gives
For the check, set \(s=\sqrt{49-x^2}\), so \(s'=-x/s\). Differentiating the answer gives
This verifies both the logarithmic expression and its coefficient.
Worked Example: Keeping the Sign on a Negative Interval
Find an antiderivative of \(\displaystyle\frac{3}{x\sqrt{4-x^2}}\) on \((-2,0)\). The sine substitution is \(x=2\sin\theta\), with \(-\pi/2<\theta<0\) on this interval. Then \(dx=2\cos\theta\,d\theta\) and \(\sqrt{4-x^2}=2\cos\theta\), so
The signed triangle relations are \(\csc\theta=2/x\) and \(\cot\theta=\sqrt{4-x^2}/x\). Both are negative here, but their difference is \(x/(2+\sqrt{4-x^2})\), also negative. The absolute value is therefore essential:
To verify, put \(s=\sqrt{4-x^2}\), with \(s'=-x/s\). Then
The derivative check applies on \((-2,0)\), where all expressions in the integrand are defined.
Equivalent Logarithmic Forms
The expression inside the logarithm can be written in several equivalent ways. For example, rationalizing gives \((a-\sqrt{a^2-x^2})/x\). Thus, wherever the expressions are defined,
These forms differ in appearance but not in value: rationalizing shows that their arguments are equal. The form with \(x\) in the numerator often makes the back-substitution especially direct. Keeping an absolute value is important on both intervals because the argument is negative when \(x<0\). A logarithm of a negative real number is not defined, while the absolute value makes the real antiderivative valid on either domain interval.
One can also read off a constant multiple immediately. For any real constant \(k\),
This follows by multiplying the integrand and its antiderivative by \(k\); it does not change the domain restrictions.
A Common Scale-Factor Mistake
A frequent error is to set \(x=a\sin\theta\) and then forget that the substitution contributes a factor of \(a\) to \(dx\), as well as a factor of \(a\) to the square root. For instance, with \(a=5\), an incorrect cancellation might be written as
The numerator contains \(5\), but the denominator contains \(25\), so the actual expression is \((1/5)\csc\theta\,d\theta\). The incorrect calculation would give \(\ln|\csc\theta-\cot\theta|\); differentiating its back-substitution yields \(5/[x\sqrt{25-x^2}]\), five times the desired integrand. The corrected antiderivative includes \(1/5\), and its derivative is \(1/[x\sqrt{25-x^2}]\).
A second safeguard is to specify the angle range before replacing the square root by \(a\cos\theta\). Without \(\cos\theta>0\), the principal square root is \(a|\cos\theta|\), not necessarily \(a\cos\theta\). The chosen range makes the simplification valid throughout the substitution.
Why the Domain and Check Matter
The original integrand has singularities at \(x=0\) and at \(x=\pm a\). The antiderivative formula is used on \((-a,0)\) or \((0,a)\), not across these excluded points as a single differentiable antiderivative. The constants of integration on the two intervals need not be the same. This domain check is not merely formal: the absolute value in the logarithm and the signed trigonometric ratios both depend on the sign of \(x\).
Differentiation provides an efficient final test. In general, with \(s=\sqrt{a^2-x^2}\), the chain rule gives \(s'=-x/s\). Therefore
Here \(s^2+x^2=a^2\), so \(s(a+s)+x^2=as+a^2=a(a+s)\). The derivative exactly recovers the original integrand, completing the check.
Check Your Understanding
Use the substitution, domain restrictions, and derivative checks from this tutorial to answer the following.
- For which real \(x\) is \(1/[x\sqrt{a^2-x^2}]\) defined when \(a>0\)?
- With \(x=a\sin\theta\), why is \(\sqrt{a^2-x^2}=a\cos\theta\) for \(-\pi/2<\theta<\pi/2\)?
- What factor remains when \(\int dx/[x\sqrt{a^2-x^2}]\) is transformed into an integral involving \(\csc\theta\)?
- Why must the logarithm in the algebraic antiderivative contain an absolute value when \(x<0\)?
- Differentiate \(\frac{2}{a}\ln\left|x/(a+\sqrt{a^2-x^2})\right|\) and identify the integrand it produces.
