Why the Derivative Suggests the Substitution
A graph \(y=f(x)\) on an interval \([a,b]\), where \(f\) is continuously differentiable, has length
The quantity inside the square root depends on the slope of the graph. A useful substitution therefore often begins with that slope: set \(u=f'(x)\). But this choice is useful only if the accompanying differential \(dx\) can also be expressed in terms of \(u\) in a manageable way. The substitution must transform the entire integrand, not just the expression under the square root.
Two patterns occur often. First, an identity may simplify the square root directly, as when \(1+\sinh^2 x=\cosh^2 x\). Second, substituting \(u=f'(x)\) may make the integral standard, provided \(f''(x)\) and the resulting differential produce a manageable expression. In either case, the interval matters: \(\sqrt{g(x)^2}=|g(x)|\), not automatically \(g(x)\).
The Standard Square-Root Integral
One common antiderivative is \(\int\sqrt{1+u^2}\,du\). Its derivation also illustrates why an algebraic rearrangement can turn an unfamiliar integral into a familiar one.
To justify the formula, write \(I=\int\sqrt{1+u^2}\,du\) and use integration by parts with the first factor \(\sqrt{1+u^2}\) and the second factor \(1\). This gives
Since \(u^2=(1+u^2)-1\), the remaining integrand is \(\sqrt{1+u^2}-1/\sqrt{1+u^2}\). Therefore
The last antiderivative is \(\ln|u+\sqrt{1+u^2}|\): differentiating the logarithm gives \(1/\sqrt{1+u^2}\). Solving the displayed equation for \(I\) proves the formula. The logarithm’s argument is positive for every real \(u\), since \(\sqrt{1+u^2}>|u|\), but the absolute-value notation is conventional.
When \(u=f'(x)\), the standard formula alone does not complete a graph-length calculation. One must still determine \(du=f''(x)\,dx\), account for any resulting factors, and transform the interval endpoints. A substitution is effective when all these pieces fit together.
Worked Example: A Hyperbolic Derivative
Worked Example: A Hyperbolic Derivative
Find the length of \(f(x)=\tfrac12\cosh(2x)\) on \(0\le x\le\tfrac12\).
Differentiate the graph function:
Thus the length integrand is \(\sqrt{1+\sinh^2(2x)}\). The identity \(1+\sinh^2 t=\cosh^2 t\) gives \(\sqrt{1+\sinh^2(2x)}=|\cosh(2x)|\). Since \(\cosh(2x)>0\), this is \(\cosh(2x)\).
Now use the derivative-based substitution \(u=\sinh(2x)\). Then \(du=2\cosh(2x)\,dx\), so \(\cosh(2x)\,dx=\tfrac12du\). At the endpoints, \(x=0\) gives \(u=0\), and \(x=\tfrac12\) gives \(u=\sinh(1)\). Therefore
For a check, an antiderivative in \(x\) is \(\tfrac12\sinh(2x)\), whose derivative is \(\cosh(2x)\). Evaluating it at the endpoints gives \(\tfrac12\sinh(1)-\tfrac12\sinh(0)=\tfrac12\sinh(1)\), as obtained by substitution.
Worked Example: A Trigonometric Derivative
Worked Example: A Trigonometric Derivative
Find the length of \(f(x)=-\ln(\cos x)\) on \([-\pi/4,\pi/4]\). On this interval \(\cos x>0\), so the function is defined and differentiable.
Its derivative is \(f'(x)=\tan x\). The identity \(1+\tan^2 x=\sec^2 x\) gives \(\sqrt{1+\tan^2 x}=|\sec x|\). Here \(\sec x>0\), so the integrand is \(\sec x\). Set \(u=\tan x\). Then \(du=\sec^2 x\,dx=(1+u^2)\,dx\), and \(\sec x=\sqrt{1+u^2}\). Consequently, \(\sec x\,dx=du/\sqrt{1+u^2}\). The transformed endpoints are \(-1\) and \(1\), so
At \(u=1\), the logarithm is \(\ln(1+\sqrt2)\). At \(u=-1\), its argument is \(\sqrt2-1\), which is positive. Thus
The last equality follows because \((1+\sqrt2)/(\sqrt2-1)=(1+\sqrt2)^2\). To check, the derivative of \(\ln|\sec x+\tan x|\) is \(\sec x\), wherever the expression is defined. Evaluating this antiderivative at the two endpoints gives \(2\ln(1+\sqrt2)\), confirming the result.
Worked Example: An Algebraic Derivative
Worked Example: An Algebraic Derivative
Find the length of \(f(x)=-\sqrt{1-x^2}\) on \([0,3/5]\). The interval lies inside \((-1,1)\), where the function is differentiable.
Differentiation gives \(f'(x)=x/\sqrt{1-x^2}\). Hence
The last equality uses \(1-x^2>0\) on the interval. To express this through the derivative, set \(u=f'(x)=x/\sqrt{1-x^2}\). Then
Also, \(1+u^2=1/(1-x^2)\), so \(du/dx=(1+u^2)^{3/2}\). It follows that \(dx=du/(1+u^2)^{3/2}\), and the length integrand becomes \(du/(1+u^2)\). The bounds transform from \(x=0\) to \(u=0\), and from \(x=3/5\) to \(u=(3/5)/(4/5)=3/4\). Therefore
For a check in the original variable, \(\frac{d}{dx}\arcsin x=1/\sqrt{1-x^2}\) on this interval. Thus the length is \(\arcsin(3/5)-\arcsin(0)=\arcsin(3/5)\). These two exact values agree: if \(\theta=\arctan(3/4)\), a right triangle with opposite side \(3\), adjacent side \(4\), and hypotenuse \(5\) gives \(\sin\theta=3/5\), hence \(\theta=\arcsin(3/5)\).
Check the Sign Before Simplifying
A frequent error is to replace \(\sqrt{1+\tan^2 x}\) by \(\sec x\) without checking its sign. The identity gives \(\sqrt{\sec^2 x}=|\sec x|\). If \(\sec x<0\), dropping the absolute value can even produce a negative answer for a length.
For example, use the graph \(f(x)=-\ln|\cos x|\) on that interval. Its derivative is \(\tan x\), and \(|\cos x|>0\) throughout. The correct length is
Indeed, \(\int\sec x\,dx=\ln|\sec x+\tan x|+C\); at the lower endpoint \(|\sec x+\tan x|=2+\sqrt3\), and at the upper endpoint it is \(1+\sqrt2\). The ratio is greater than one, so its logarithm is positive.
A Reliable Substitution Checklist
Before committing to a substitution, verify that it simplifies the complete length integral. In particular, if \(u=f'(x)\), then \(du=f''(x)\,dx\); a factor of \(f''(x)\) must not be omitted. For a definite integral, transform both endpoints or return to \(x\) before evaluating. Also check whether the interval crosses a point where the expression used to simplify the square root changes sign or is undefined.
Begin with \(\sqrt{1+[f'(x)]^2}\) and identify its algebraic, trigonometric, or hyperbolic structure.
Use \(\sqrt{g(x)^2}=|g(x)|\), and determine whether the graph and substitution are defined throughout the interval.
Compute the differential and, for a definite integral, the new bounds. Simplifying only the square root is not enough.
Differentiate the proposed antiderivative or compare with the original-variable integral to verify the result.
These patterns matter because exact curve lengths depend not merely on recognizing a familiar expression, but on matching it to the graph’s derivative and the interval. When a substitution does not simplify the full integral, it is not a useful substitution, even if it makes the square root look simpler. The standard integral of \(\sqrt{1+u^2}\) is one available tool; choosing when it applies is part of the calculation.
Check Your Understanding
For each question, state any interval or sign condition needed to justify your reasoning.
- Why does \(u=f'(x)\) require transforming the differential as well as the expression inside the square root?
- State an antiderivative of \(\sqrt{1+u^2}\), and name the integration step used to derive it.
- On what intervals can \(\sqrt{1+\tan^2 x}\) be replaced by \(\sec x\) without an absolute value?
- For \(f'(x)=\sinh(2x)\), what substitution transforms the length integrand and differential together?
- For \(f'(x)=x/\sqrt{1-x^2}\), find \(du/dx\) when \(u=f'(x)\), and state the transformed integrand.
