Definite Integrals Of Cotangent And Cosecant Powers

Definite Integrals Require Attention to the Interval

A definite integral of cotangent and cosecant powers is not determined by an antiderivative alone. The interval matters: both \(\cot x\) and \(\csc x\) are undefined where \(\sin x=0\), at \(x=k\pi\) for integers \(k\). Before evaluating, check whether any such point is an endpoint or lies inside the interval.

Throughout this tutorial, \(m\) and \(n\) are nonnegative integers, and the integrand is \(\cot^m x\,\csc^n x\). When the interval contains no point \(k\pi\), the integrand is continuous there and the Fundamental Theorem of Calculus applies: if \(F'(x)=\cot^m x\,\csc^n x\), then the integral from \(a\) to \(b\) is \(F(b)-F(a)\).

For positive \(m+n\), the singularities at \(k\pi\) are not removable. Near \(x=k\pi\), \(\sin x\) behaves like \(x-k\pi\), while \(\cos x\) approaches \(1\) or \(-1\). Thus the magnitude of \(\cot^m x\,\csc^n x\) behaves like a nonzero constant times \(|x-k\pi|^{-(m+n)}\). Since \(m+n\geq 1\), the corresponding one-sided integral diverges. A principal value based on cancellation is a different notion and does not make the ordinary improper integral converge.

Evaluate First by Substitution When It Fits

The derivative identities \(\frac{d}{dx}(\cot x)=-\csc^2 x\) and \(\csc^2 x=1+\cot^2 x\) make a cotangent substitution effective when a factor of \(\csc^2 x\) is available. On an interval where the integrand is continuous, set \(u=\cot x\), so \(du=-\csc^2 x\,dx\), and transform the limits along with the integrand.

Definite Reduction Formulas

When a direct substitution does not finish the integral, reduction formulas replace a higher power by a lower one and an explicit endpoint term. The endpoint term must be evaluated at both limits; the recurrence does not discard it.

Write \(I_n(a,b)=\int_a^b\csc^n x\,dx\) on an interval avoiding all \(k\pi\). For integers \(n\geq 2\), integration by parts gives the cosecant-power recurrence

It reduces even powers to \(I_0(a,b)=b-a\), and odd powers to \(I_1\). The base case \(I_1\), on any interval avoiding the poles, is given by the antiderivative \(\ln|\csc x-\cot x|\). The recurrence requires \(n\geq2\); it should not be applied to \(n=1\).

There is also a reduction formula for mixed powers. Let \(J_{m,n}(a,b)=\int_a^b\cot^m x\,\csc^n x\,dx\), with \(m\geq1\), \(m+n>1\), and an interval avoiding the poles. Then

For \(m=1\), the last term has coefficient zero, so the formula evaluates the integral directly. For \(m\geq2\), it lowers the cotangent exponent by two. The stated conditions matter: when \(m+n=1\), the denominator vanishes and this recurrence does not apply.

Why the Mixed Recurrence Works

The recurrence follows by differentiating a product, then using \(\csc^2 x=1+\cot^2 x\). On an interval without poles, all the functions involved are differentiable, so the product rule and the Fundamental Theorem of Calculus apply.

Check Endpoint Behavior Before Using an Antiderivative

An antiderivative formula does not establish that an improper integral converges. For example, the integral of \(\csc x\) from \(0\) to \(\pi/2\) has a singular endpoint. Using \(\int\csc x\,dx=\ln|\csc x-\cot x|\), for \(0<\varepsilon<\pi/2\) the truncated integral is

Here \(\csc\varepsilon-\cot\varepsilon=\tan(\varepsilon/2)\), which is positive and tends to zero as \(\varepsilon\to0^+\). Therefore the logarithm tends to \(-\infty\), and the integral tends to \(+\infty\). The improper integral diverges; the antiderivative does not give it a finite value.

A Reliable Evaluation Checklist

For a definite cotangent-cosecant power integral, first identify the interval and test it for poles \(k\pi\). If the interval avoids them, choose an applicable substitution or reduction formula, keep all endpoint terms, and reduce only to a valid base case. If the interval has a singular endpoint or an interior pole, evaluate the required one-sided limits before treating the result as an ordinary finite integral.

These checks are part of the evaluation, not optional bookkeeping. They distinguish a valid finite definite integral from an expression whose antiderivative is usable only on pole-free intervals, or whose improper limit does not exist finitely.