Why simplify before integrating?
For a parametric curve \(x=x(t)\), \(y=y(t)\) on \(a\le t\le b\), its arc length is the integral of its speed:
When the coordinate functions involve sine and cosine, the expression under the square root may look more complicated than it really is. Expanding and combining terms can reveal a constant speed, an absolute sine or cosine, or a standard integral that does not have an elementary antiderivative. The key is to simplify the squared speed first, then take its nonnegative square root.
Start with the derivatives, not the coordinates
Suppose the coordinates are combinations of sine and cosine with the same frequency \(k\):
Here \(a,b,c,d\), and \(k\) are constants. Differentiating gives
To simplify the speed, square these expressions and add them. Cross terms may cancel, and the identity \(\sin^2 u+\cos^2 u=1\) may turn a variable expression into a constant. Do not replace a square root of a square with its contents automatically: \(\sqrt{q(t)^2}=|q(t)|\), not necessarily \(q(t)\).
Worked Example: Cross terms produce constant speed
Let \(x(t)=2\cos t+2\sin t\) and \(y(t)=2\cos t-2\sin t\). Find the speed and the arc length for \(0\le t\le \frac{\pi}{2}\).
Differentiate each coordinate:
Write \(s=\sin t\) and \(c=\cos t\) just for the expansion. Then
Therefore the speed is \(v(t)=\sqrt{8}=2\sqrt{2}\). It is constant, so the arc length is
Check: differentiating the accumulated-length function \(S(t)=2\sqrt{2}\,t\) gives \(S'(t)=2\sqrt{2}\), the speed found above. The length is positive, as it must be.
A general reduction for one frequency
For coordinate functions made from \(\sin(kt)\) and \(\cos(kt)\), the squared speed is a quadratic expression in \(\sin(kt)\) and \(\cos(kt)\). It can include a cross term. A useful way to understand the simplification is to regard that expression as a quadratic form. In the result below, a change in phase shifts the angle but does not change which types of antiderivative are possible.
Here is the reduction and the reason it works. Put \(w=(\cos(kt),\sin(kt))^\top\), and define the matrix \(M=\begin{pmatrix}b&-a\\ d&-c\end{pmatrix}\). The velocity vector is \(kMw\), so its squared length is \(k^2 w^\top M^\top M w\). The matrix \(M^\top M\) is symmetric and positive semidefinite: for every vector \(z\), \(z^\top M^\top Mz=(Mz)^\top(Mz)\ge0\). A symmetric matrix has orthogonal eigenvectors, so rotating the angle coordinates to those eigenvector directions expresses this quadratic form using its nonnegative eigenvalues \(\lambda_1,\lambda_2\). A rotation of the unit vector \(w\) is just a phase shift, yielding the stated expression.
This reduction also explains the antiderivative test. If \(\lambda_1=\lambda_2=\lambda\), then the squared trigonometric terms add to \(1\), and the speed is \(|k|\sqrt{\lambda}\). If one eigenvalue is zero, taking the square root leaves an absolute sine or cosine, which can be integrated on intervals where its sign is fixed. If both are positive but unequal, let \(\lambda_{\max}\) be the larger eigenvalue and \(\lambda_{\min}\) the smaller. The speed has the form
The integral of \(\sqrt{1-m\sin^2u}\) is a complete or incomplete elliptic integral of the second kind, depending on the bounds. For \(0<m<1\), this integrand has no elementary antiderivative in general. This classification is useful: simplifying the trigonometry can identify when an exact elementary calculation is available, and when it is not.
When the square root becomes an absolute value
A zero eigenvalue, or a direct factorization of the squared speed, can produce \(\sqrt{C\sin^2 u}=\sqrt{C}\,|\sin u|\) for \(C\ge0\). The absolute value matters if the parameter interval crosses a zero or a change of sign. On any interval where \(\cos u\) has fixed sign, for example, an antiderivative of \(|\cos u|\) is \(\operatorname{sgn}(\cos u)\sin u\). Its derivative is \(\operatorname{sgn}(\cos u)\cos u=|\cos u|\). If the sign changes within the interval, split the integral there.
Worked Example: A rank-one velocity gives an absolute sine
Let \(x(t)=3\cos t\) and \(y(t)=4\cos t\), for \(\pi\le t\le2\pi\). Find the speed and arc length.
The derivatives are \(x'(t)=-3\sin t\) and \(y'(t)=-4\sin t\). Thus
On \([\pi,2\pi]\), \(\sin t\le0\), so \(|\sin t|=-\sin t\). Therefore
Check: the derivative of \(5\cos t\) is \(-5\sin t\), which equals the speed on this interval. The endpoint evaluation gives a positive length of \(10\).
Unequal axes and a non-elementary integral
An ellipse parametrized with different horizontal and vertical scale factors is a common source of an integral that cannot be simplified to elementary functions. The trigonometric identity still gives a useful reduced form, even though it does not provide an elementary antiderivative.
Worked Example: Recognizing an elliptic arc-length integral
For \(x(t)=3\cos t\) and \(y(t)=4\sin t\), find the speed on \(0\le t\le\frac{\pi}{2}\), simplify it, and decide whether it has an elementary antiderivative.
Differentiate:
Then simplify the squared speed:
Thus
The parameter is \(m=\frac{7}{16}\), which satisfies \(0<m<1\). This is the unequal, positive-eigenvalue case of the theorem, so this integrand does not have an elementary antiderivative in general. The arc length is exactly represented by the definite integral
As a check on the speed formula, at \(t=0\) it gives \(4\), matching \(\sqrt{0^2+4^2}=4\). At \(t=\frac{\pi}{2}\) it gives \(3\), matching \(\sqrt{(-3)^2+0^2}=3\). Those checks confirm the simplification, though endpoint checks alone do not establish whether an antiderivative is elementary.
Common mistake: dropping the absolute value
A frequent error is to take the square root of \(9\sin^2t\) as \(3\sin t\). The correct identity is \(\sqrt{9\sin^2t}=3|\sin t|\). For example, on \([\pi,2\pi]\), \(\sin t\le0\), so \(3\sin t\) would be negative even though speed cannot be negative. Using that incorrect expression would give
which cannot be a length. The correction is \(3|\sin t|=-3\sin t\) on this interval, giving
So the reliable sequence is: calculate the squared speed, simplify it, take the nonnegative square root, and only then use the parameter interval to resolve any absolute values.
How to choose the next step
Once the speed is simplified, inspect its form before attempting an integration method. A constant can be integrated directly. A constant multiple of an absolute sine or cosine can be integrated by splitting at its zeros if needed. An expression of the form \(\sqrt{A-B\sin^2t}\), with \(A>B>0\), is generally elliptic rather than elementary. Recognizing this distinction prevents misleading algebraic attempts to produce an elementary formula where none exists.
Arc length uses the velocity components, not the original coordinate functions.
Combine cross terms and use trigonometric identities before taking the square root.
Preserve absolute values and split the interval at sign changes when necessary.
Distinguish constant or absolute-trigonometric speed from the unequal positive-axis elliptic form.
Check Your Understanding
For each question, simplify carefully and use the parameter interval when determining the speed.
- If \(x(t)=5\cos t\) and \(y(t)=5\sin t\), what is the speed? Is its antiderivative elementary?
- Why does \(\sqrt{16\sin^2t}\) equal \(4|\sin t|\), rather than always \(4\sin t\)?
- For \(x(t)=2\cos t\), \(y(t)=6\sin t\), what form does the speed take after using \(\sin^2t+\cos^2t=1\)?
- What parameter value \(m\) appears in the reduced integrand for the curve in question 3, and what does it indicate about elementary antiderivatives?
- What must be done if an absolute sine or cosine changes sign inside the parameter interval?
