Deriving The Cotangent-Power Reduction Formula

Why Reduce a Cotangent Power?

Integrals of high powers of cotangent can look increasingly complicated as the exponent grows. The useful observation is that the identity $\cot^2 x=\csc^2 x-1$ lets us relate one power to a power two units lower. Repeatedly applying that relation reduces the problem to two simple base cases.

The derivation matters as much as the resulting formula. In particular, the recurrence is not a rule for every exponent: it applies to integer powers $n\ge 2$, and its reduction eventually reaches either the zeroth-power or first-power integral. We will establish those base cases and check that the recurrence produces correct antiderivatives.

Definition: For a nonnegative integer $n$, write $I_n=\int \cot^n x\,dx$ for an antiderivative of $\cot^n x$. These antiderivatives are considered on an interval where $\sin x\ne 0$, so that $\cot x$ is defined and continuous. An arbitrary constant of integration is understood.

The Cotangent Identity and the Recurrence

The basic trigonometric identity follows from $\csc^2 x=1+\cot^2 x$, so $\cot^2 x=\csc^2 x-1$. To derive the reduction formula, split a power into a factor of $\cot^2 x$ and the remaining factor. The integration-by-parts step uses the differential of $\cot x$: since $\frac{d}{dx}\cot x=-\csc^2 x$, the factor $\csc^2 x\,dx$ can be expressed as $-d(\cot x)$.

Theorem: For an integer $n\ge 2$, an antiderivative of $\cot^n x$ satisfies $$ I_n=-\frac{\cot^{n-1}x}{n-1}-I_{n-2}. $$ Equivalently, $$ \int \cot^n x\,dx = -\frac{\cot^{n-1}x}{n-1} -\int \cot^{n-2}x\,dx. $$ The formula is used on an interval where $\sin x\ne 0$.

Here is the integration-by-parts derivation. First, when $n=2$, the identity gives $\cot^2 x=\csc^2 x-1$, and direct integration yields $I_2=-\cot x-x$. Now take an integer $n\ge 3$. Start by separating two powers:

I_n = \int \cot^{n-2}x\,\cot^2 x\,dx = \int \cot^{n-2}x\,\csc^2 x\,dx-I_{n-2}.

In the first integral, use $d(\cot x)=-\csc^2 x\,dx$. Integration by parts, in the form $\int f\,(-dg)=-fg+\int g\,df$, with $f=\cot^{n-2}x$ and $g=\cot x$, gives

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

Since $d(\cot x)=-\csc^2 x\,dx$, and $\csc^2 x=\cot^2 x+1$, the remaining integral becomes

\int \cot^{n-2}x\,d(\cot x) = -\int \cot^{n-2}x\,(\cot^2 x+1)\,dx = -I_n-I_{n-2}.

Substituting this expression into the preceding equations and collecting the $I_n$ terms gives

I_n = -\cot^{n-1}x -(n-2)(I_n+I_{n-2}) -I_{n-2}, $$ $$ (n-1)I_n = -\cot^{n-1}x-(n-1)I_{n-2}.

Dividing by $n-1$, which is nonzero for $n\ge 3$, proves the stated recurrence for these powers. The direct calculation for $n=2$ gives the same formula at that endpoint. The derivation shows why each reduction lowers the exponent by two: the cotangent-square identity separates the integrand into a cosecant-square term and a term with two fewer cotangent factors.

The Base Cases

The recurrence eventually reaches $I_0$ or $I_1$, depending on whether the original exponent is even or odd. These two antiderivatives are therefore essential parts of the method.

Base cases: On an interval where $\sin x\ne 0$, $$ I_0=\int 1\,dx=x+C, \qquad I_1=\int \cot x\,dx=\ln|\sin x|+C. $$ The logarithm is defined because $\sin x\ne 0$ on the interval.

For the first case, differentiating $x$ gives $1$. For the second, the derivative of $\ln|\sin x|$ is $\frac{\cos x}{\sin x}=\cot x$ wherever $\sin x\ne 0$. The absolute value is important because $\sin x$ need not be positive on the chosen interval.

Using the Formula and Checking the Results

Worked Example: Reduce the Second Power

Use the cotangent identity to find an antiderivative of $\cot^2 x$. Since $\cot^2 x=\csc^2 x-1$, integrate each term:

\int \cot^2 x\,dx = \int \csc^2 x\,dx-\int 1\,dx = -\cot x-x+C.

Check by differentiation. The derivative of $-\cot x-x$ is $\csc^2 x-1$, which equals $\cot^2 x$. Thus the antiderivative is correct on any interval where $\sin x\ne 0$.

Worked Example: Reduce the Third Power

Apply the recurrence with $n=3$, then use the first-power base case:

I_3 = -\frac{\cot^2 x}{2}-I_1 = -\frac{\cot^2 x}{2}-\ln|\sin x|+C.

To check, differentiate each term. Using $\frac{d}{dx}\cot x=-\csc^2 x$, we obtain

\frac{d}{dx}\left(-\frac{\cot^2 x}{2}-\ln|\sin x|\right) = \cot x\,\csc^2 x-\cot x = \cot x(\csc^2 x-1) = \cot^3 x.

The last equality uses $\csc^2 x-1=\cot^2 x$, so the derivative matches the integrand.

Worked Example: Reduce the Fourth Power

For $n=4$, the recurrence reduces the integral to $I_2$. Substitute the second-power result:

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

Differentiate to check:

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

Factor the first two terms and use $\csc^2 x=\cot^2 x+1$:

\cot^2 x\,\csc^2 x-\csc^2 x+1 = (\cot^2 x-1)\csc^2 x+1 = (\cot^2 x-1)(\cot^2 x+1)+1 = \cot^4 x.

The derivative is the required fourth power, confirming the result.

How to Apply the Recurrence Reliably

Each use lowers the exponent by two. Consequently, repeated reduction from an even exponent reaches $I_2$ and then $I_0$; repeated reduction from an odd exponent reaches $I_3$ and then $I_1$. The recurrence itself can also reduce $I_3$ directly to $I_1$. Keeping the two base cases visible prevents a common mistake: stopping at a reduced integral without evaluating it.

Check the exponent.
The recurrence applies to integer powers $n\ge 2$, and the integral must be considered on an interval where cotangent is defined.

Reduce by two.
Replace $I_n$ with $-\cot^{n-1}x/(n-1)-I_{n-2}$, preserving the minus sign before the lower-power integral.

Use a base case.
Evaluate the remaining $I_0$ or $I_1$, including its constant of integration.

Differentiate to verify.
Check that the derivative of the proposed antiderivative equals the original cotangent power.

A Common Sign Error

The sign in the recurrence is easy to reverse. A tempting but incorrect calculation is to write $I_2=\cot x-x$, as if $\int\csc^2 x\,dx=\cot x$. In fact, $\frac{d}{dx}\cot x=-\csc^2 x$, so the correct antiderivative of $\csc^2 x$ is $-\cot x$. Therefore,

I_2=\int(\csc^2 x-1)\,dx=-\cot x-x+C,

not $\cot x-x+C$. Differentiating the incorrect expression gives $-\csc^2 x-1$, which is not $\cot^2 x=\csc^2 x-1$. Checking by differentiation catches the sign error immediately.

A related limitation is that the recurrence is not applied at $n=1$: its derivation requires splitting off two powers, and the displayed formula would involve division by $n-1=0$. The first power is instead handled by its logarithmic base case. This distinction keeps the reduction method valid at every step.

Why the Recurrence Is Useful

The recurrence turns a family of integrals into a systematic reduction rather than a fresh calculation for every exponent. Its proof also illustrates the role of integration by parts: the derivative of cotangent introduces $\csc^2 x$, and the identity $\csc^2 x=\cot^2 x+1$ converts that factor back into cotangent powers. The resulting relation connects an integral to one with a smaller exponent, while the base cases finish the process.

Check Your Understanding

Answer these questions, and differentiate any proposed antiderivative as a check.

State the cotangent-power reduction formula and its condition on $n$.
Why does the recurrence lower the power by two rather than by one?
What are the antiderivatives $I_0$ and $I_1$, and where are they valid?
Use the recurrence to express $I_5$ in terms of $I_3$, then reduce it to the base case.
Identify the sign error in the claim $\int \csc^2 x\,dx=\cot x+C$.

Deriving The Cotangent-Power Reduction Formula

What You'll Learn

Why Reduce a Cotangent Power?

The Cotangent Identity and the Recurrence

The Base Cases