Evaluating Higher Cotangent Powers By Recurrence

Reducing Cotangent Powers

Integrals of high powers of cotangent can look increasingly complicated as the exponent grows. The cotangent-power recurrence gives a systematic way to reduce such an integral to one with an exponent two units smaller. Repeating that step eventually reaches a simple base integral, so the method works for both even and odd positive integer powers.

The main practical challenge is keeping track of the alternating signs and stopping at the correct base case. An even power eventually reduces to the integral of \(1\); an odd power eventually reduces to the integral of \(\cot x\), which introduces a logarithm. In each case, differentiating the proposed answer provides a direct check.

The Recurrence and Its Base Cases

For a nonnegative integer \(n\), write \(I_n\) for an antiderivative of \(\cot^n x\). The formulas below apply on any interval on which \(\sin x\ne 0\), so that \(\cot x\) is defined throughout the interval. As usual, antiderivatives on the same interval may differ by a constant.

Definition: For each nonnegative integer \(n\), let \(I_n=\int \cot^n x\,dx\). The base cases are \(I_0=x\) and \(I_1=\ln|\sin x|\), up to an additive constant. For each integer \(n\ge 2\), the cotangent-power recurrence is \(I_n=-\frac{\cot^{n-1}x}{n-1}-I_{n-2}\).

The recurrence lowers the exponent by two, not by one. This is why the parity of the starting exponent matters: an even exponent reaches \(I_0\), while an odd exponent reaches \(I_1\). The base case \(I_0=x\) follows from \(\cot^0x=1\). For the other base case, \(\frac{d}{dx}\ln|\sin x|=\frac{\cos x}{\sin x}=\cot x\) wherever \(\sin x\ne0\).

The recurrence follows from the identity \(\csc^2x=1+\cot^2x\) and the derivative \(\frac{d}{dx}\cot x=-\csc^2x\). The short derivation also explains the minus sign in the formula.

\begin{aligned} \cot^n x &=\cot^{n-2}x\bigl(\csc^2x-1\bigr),\\ I_n &=\int \cot^{n-2}x\,\csc^2x\,dx-I_{n-2}\\ &=-\frac{\cot^{n-1}x}{n-1}-I_{n-2}. \end{aligned}

For the last equality, set \(u=\cot x\), so \(du=-\csc^2x\,dx\). Then the first integral becomes \(-\int u^{n-2}\,du=-u^{n-1}/(n-1)\), which is the displayed cotangent term. This substitution is valid for \(n\ge2\), and the exponent \(n-2\) is then a nonnegative integer.

Why the method terminates: Each recurrence step replaces \(n\) by \(n-2\). After finitely many steps, the exponent is either \(0\) or \(1\), where the corresponding base antiderivative is known.

Worked Examples

In each example, apply the recurrence repeatedly until reaching a base case. Keep the negative sign before the lower-power integral: it affects the signs of every later term.

Worked Example: Integrating \(\cot^2x\)

Apply the recurrence with \(n=2\). Its lower-power term is \(I_0=x\):

\begin{aligned} I_2 &=-\frac{\cot^{1}x}{1}-I_0\\ &=-\cot x-x. \end{aligned}

Thus \(\int\cot^2x\,dx=-\cot x-x+C\). To check, differentiate the expression without the constant. Since \(\frac{d}{dx}(-\cot x)=\csc^2x\) and \(\frac{d}{dx}(-x)=-1\),

\frac{d}{dx}(-\cot x-x) =\csc^2x-1 =(1+\cot^2x)-1 =\cot^2x.

The derivative equals the integrand, as required.

Worked Example: Integrating \(\cot^4x\)

First reduce \(I_4\) to \(I_2\), then use the result for the lower power:

\begin{aligned} I_4 &=-\frac{\cot^3x}{3}-I_2\\ &=-\frac{\cot^3x}{3}-(-\cot x-x)\\ &=-\frac{\cot^3x}{3}+\cot x+x. \end{aligned}

Therefore \(\int\cot^4x\,dx=-\frac{\cot^3x}{3}+\cot x+x+C\). Differentiate every term to check:

\begin{aligned} \frac{d}{dx}\left(-\frac{\cot^3x}{3}+\cot x+x\right) &=\cot^2x\,\csc^2x-\csc^2x+1\\ &=\cot^2x(1+\cot^2x)-(1+\cot^2x)+1\\ &=\cot^2x+\cot^4x-1-\cot^2x+1\\ &=\cot^4x. \end{aligned}

The cancellation confirms the answer.

Worked Example: Integrating \(\cot^5x\)

This odd power descends to \(I_1\). Apply the recurrence twice and then use the logarithmic base case:

\begin{aligned} I_5 &=-\frac{\cot^4x}{4}-I_3,\\ I_3 &=-\frac{\cot^2x}{2}-I_1\\ &=-\frac{\cot^2x}{2}-\ln|\sin x|. \end{aligned}

Substitute the expression for \(I_3\) into the first line:

\begin{aligned} I_5 &=-\frac{\cot^4x}{4} -\left(-\frac{\cot^2x}{2}-\ln|\sin x|\right)\\ &=-\frac{\cot^4x}{4}+\frac{\cot^2x}{2}+\ln|\sin x|. \end{aligned}

Differentiate to check, using \(\frac{d}{dx}\ln|\sin x|=\cot x\):

\begin{aligned} \frac{d}{dx}\left(-\frac{\cot^4x}{4} +\frac{\cot^2x}{2}+\ln|\sin x|\right) &=\cot^3x\,\csc^2x-\cot x\,\csc^2x+\cot x\\ &=\cot^3x(1+\cot^2x)-\cot x(1+\cot^2x)+\cot x\\ &=\cot^3x+\cot^5x-\cot x-\cot^3x+\cot x\\ &=\cot^5x. \end{aligned}

The logarithm is needed because the recurrence reaches the \(I_1\) base case.

Worked Example: Integrating \(\cot^6x\)

For an even power, continue down to \(I_0\). First obtain \(I_4\) from the recurrence, and then use the expression for \(I_4\):

\begin{aligned} I_6 &=-\frac{\cot^5x}{5}-I_4\\ &=-\frac{\cot^5x}{5} -\left(-\frac{\cot^3x}{3}+\cot x+x\right)\\ &=-\frac{\cot^5x}{5}+\frac{\cot^3x}{3}-\cot x-x. \end{aligned}

To verify the result, differentiate and rewrite \(\csc^2x\) as \(1+\cot^2x\):

\begin{aligned} \frac{d}{dx}\left(-\frac{\cot^5x}{5} +\frac{\cot^3x}{3}-\cot x-x\right) &=\cot^4x\,\csc^2x-\cot^2x\,\csc^2x+\csc^2x-1\\ &=\cot^4x(1+\cot^2x)-\cot^2x(1+\cot^2x)\\ &\qquad +(1+\cot^2x)-1\\ &=\cot^4x+\cot^6x-\cot^2x-\cot^4x+\cot^2x\\ &=\cot^6x. \end{aligned}

All lower powers cancel, leaving exactly the original integrand.

A Reliable Way to Apply the Recurrence

When the exponent is larger, it is useful to organize the calculation before simplifying. Write each reduction in terms of the next lower integral, and stop when the exponent reaches the appropriate base case. Only then substitute the base value and collect the terms.

Identify the exponent.
Check that the power \(n\) is an integer with \(n\ge2\) before applying the recurrence.

Reduce by two.
Replace \(I_n\) with \(-\frac{\cot^{n-1}x}{n-1}-I_{n-2}\), including both the denominator and the minus sign.

Continue to a base case.
Use \(I_0=x\) for an even starting power or \(I_1=\ln|\sin x|\) for an odd starting power.

Substitute and simplify.
Insert the lower-power results carefully, then combine like terms and include one arbitrary constant.

Differentiate to check.
Use \(\frac{d}{dx}\cot x=-\csc^2x\) and \(\csc^2x=1+\cot^2x\) to verify that the derivative is the original power.

Patterns by Parity

The first few reductions reveal what to expect, without replacing the need to apply the recurrence correctly. For even powers, the final base term contributes \(x\). For odd powers, the final base term contributes \(\ln|\sin x|\). Intermediate cotangent powers alternate in sign as the recurrence is applied repeatedly.

Starting power	Base case reached	Type of final expression
Even \(n\ge2\)	\(I_0=x\)	Cotangent powers together with a linear term
Odd \(n\ge3\)	\(I_1=\ln\|\sin x\|\)	Cotangent powers together with a logarithm

This pattern is useful as a check on the form of an answer. It does not determine every coefficient by itself; those coefficients come from the recurrence. Also, the recurrence applies to a single cotangent power. It does not by itself evaluate an integral with additional cosecant factors or other trigonometric factors.

Common Sign Error

A frequent mistake is to substitute a lower-power antiderivative but forget that the recurrence subtracts the entire lower-power integral. For example, the correct step for \(I_4\) is

I_4=-\frac{\cot^3x}{3}-I_2 =-\frac{\cot^3x}{3}-(-\cot x-x) =-\frac{\cot^3x}{3}+\cot x+x.

An incorrect calculation might remove the parentheses without changing the signs:

-\frac{\cot^3x}{3}-(-\cot x-x) \ \not=\ -\frac{\cot^3x}{3}-\cot x-x.

Subtracting a negative term makes that term positive: \(-(-\cot x-x)=\cot x+x\). The incorrect expression also fails the derivative check. Its derivative is \(\cot^2x\,\csc^2x+\csc^2x-1\), which simplifies to \(\cot^4x+2\cot^2x\), not \(\cot^4x\). Differentiation therefore detects the sign error immediately.

Why the Recurrence Is Useful

Directly transforming a high cotangent power into other trigonometric functions can create expressions that are harder to integrate. The recurrence instead reduces the exponent in a predictable sequence and keeps the calculation in powers of cotangent. Because each step is explicit, it is straightforward to see which base case will occur and to check the final result.

Keep the domain in view as well. The integrand \(\cot^n x\) is defined only where \(\sin x\ne0\). An antiderivative formula is understood on an interval that does not cross a point where \(\sin x=0\). In particular, the logarithmic expression \(\ln|\sin x|\) is defined on such intervals, and its derivative there is \(\cot x\).

Takeaway: To integrate an integer power of cotangent, reduce the exponent by two at each step, preserve the minus sign before the lower-power integral, use \(I_0=x\) or \(I_1=\ln|\sin x|\) at the end, and differentiate the result to check it.

Check Your Understanding

Use the recurrence and the base cases to answer these questions. Check any antiderivative you obtain by differentiating it.

Which base case is reached when the starting exponent is \(8\), and which is reached when it is \(7\)?
Use one recurrence step to express \(I_3\) in terms of \(I_1\), then state the resulting antiderivative.
Use the recurrence to evaluate \(\int\cot^4x\,dx\). Where does the sign of the \(\cot x\) term come from?
Why must an interval of integration avoid points where \(\sin x=0\)?
Differentiate \(-\frac{\cot^3x}{3}+\cot x+x\) and simplify to verify the corresponding integrand.

Evaluating Higher Cotangent Powers By Recurrence

What You'll Learn

Reducing Cotangent Powers

The Recurrence and Its Base Cases