Choosing A Substitution For Mixed Cotangent-Cosecant Powers

Why the Exponents Guide the Substitution

Mixed powers of cotangent and cosecant can look less familiar than the corresponding tangent–secant integrals, but their structure is governed by two simple derivatives:

\frac{d}{dx}\cot x=-\csc^2 x, \qquad \frac{d}{dx}\csc x=-\csc x\cot x.

Each derivative suggests a factor to set aside. A factor of \(\csc^2 x\,dx\) pairs with \(u=\cot x\); a factor of \(\csc x\cot x\,dx\) pairs with \(u=\csc x\). The exponents tell us whether the integrand contains enough of the right factors, and whether the remaining powers can be rewritten as a polynomial in \(u\).

In this tutorial, \(m\) and \(n\) are nonnegative integers, and we consider integrals of the form \(\int \cot^m x\,\csc^n x\,dx\). Antiderivatives are understood on intervals where the integrand is defined, so in particular \(\sin x\ne 0\). The direct parity rules below are useful when the required derivative factors are present.

Definition: An integer exponent is even if it is divisible by \(2\), and odd if it leaves remainder \(1\) on division by \(2\). For these mixed-power integrals, an even cosecant exponent provides a pair of cosecant factors, while an odd cotangent exponent provides a cotangent factor to pair with a cosecant factor.

The Two Main Parity Patterns

Pattern 1: even cosecant exponent. If \(n\) is even and \(n\ge 2\), separate one factor \(\csc^2 x\,dx\) and let \(u=\cot x\). The remaining cosecant exponent \(n-2\) is also even. The identity \(\csc^2 x=1+\cot^2 x\) then expresses the remaining cosecant power in terms of \(u\).

\csc^2 x=1+\cot^2 x, \qquad du=-\csc^2 x\,dx.

For example, the general transformation in this case is

\int \cot^m x\,\csc^n x\,dx = -\int u^m(1+u^2)^{(n-2)/2}\,du, \qquad u=\cot x.

Because \((n-2)/2\) is a nonnegative integer, the transformed integrand is a polynomial when \(m\) is a nonnegative integer. Expanding and integrating that polynomial gives an antiderivative.

Pattern 2: odd cotangent exponent. If \(m\) is odd and \(n\ge 1\), separate one factor \(\cot x\csc x\,dx\) and let \(u=\csc x\). The remaining cotangent exponent \(m-1\) is even. Use \(\cot^2 x=\csc^2 x-1\) to rewrite it in terms of \(u\).

\cot^2 x=\csc^2 x-1, \qquad du=-\csc x\cot x\,dx.

This gives

\int \cot^m x\,\csc^n x\,dx = -\int (u^2-1)^{(m-1)/2}u^{n-1}\,du, \qquad u=\csc x.

Here \((m-1)/2\) is a nonnegative integer, so the transformed integrand is again a polynomial for nonnegative integer \(n\). If both conditions hold—\(n\) even and \(m\) odd—either substitution is available. It can be useful to choose whichever produces the simpler polynomial.

Why These Substitutions Work

The parity rules are consequences of matching a derivative factor and then applying a Pythagorean identity. The following derivation verifies the even-cosecant pattern completely. Let \(n=2k\), where \(k\ge 1\), and take \(u=\cot x\). Then

\begin{aligned} \int \cot^m x\,\csc^{2k}x\,dx &=\int \cot^m x\,\csc^{2k-2}x\, \bigl(\csc^2 x\,dx\bigr)\\ &=-\int u^m(1+u^2)^{k-1}\,du. \end{aligned}

The first line simply separates two of the \(2k\) cosecant factors. In the second line, \(\csc^2 x=1+\cot^2 x\) converts the unseparated factors, and \(\csc^2 x\,dx=-du\) replaces the differential. Since \(k-1\) is a nonnegative integer, the binomial expansion of \((1+u^2)^{k-1}\) has finitely many terms. Thus the substitution reduces the original integral to polynomial integration.

For the odd-cotangent pattern, write \(m=2r+1\), where \(r\ge 0\), and take \(u=\csc x\). Separating the factor \(\cot x\csc x\,dx\) gives

\begin{aligned} \int \cot^{2r+1}x\,\csc^n x\,dx &=\int \cot^{2r}x\,\csc^{n-1}x\, \bigl(\cot x\csc x\,dx\bigr)\\ &=-\int (u^2-1)^r u^{n-1}\,du. \end{aligned}

The minus sign in both formulas is essential: each substitution has a negative derivative. These derivations also explain the limits of the rules. If \(n\) is odd and \(m\) is even, neither specified factor is automatically available in the required parity pattern. In that case, do not force one of these direct substitutions; another method, such as an appropriate reduction formula, may be needed.

Worked Examples

Worked Example: Even Cosecant Exponent

Evaluate \(\int \cot^4 x\,\csc^6 x\,dx\).

The cosecant exponent \(6\) is even, so use \(u=\cot x\) and reserve \(\csc^2 x\,dx=-du\). The remaining cosecant power is \(\csc^4 x\):

\begin{aligned} \int \cot^4 x\,\csc^6 x\,dx &=\int \cot^4 x\,\csc^4 x\, \bigl(\csc^2 x\,dx\bigr)\\ &=-\int u^4(1+u^2)^2\,du\\ &=-\int (u^4+2u^6+u^8)\,du\\ &=-\frac{u^5}{5}-\frac{2u^7}{7}-\frac{u^9}{9}+C\\ &=-\frac{\cot^5 x}{5} -\frac{2\cot^7 x}{7} -\frac{\cot^9 x}{9}+C. \end{aligned}

To check, differentiate the result. Since \(\frac{d}{dx}\cot x=-\csc^2 x\), the three terms give, respectively, \(\cot^4 x\csc^2 x\), \(2\cot^6 x\csc^2 x\), and \(\cot^8 x\csc^2 x\). Their sum is

\cot^4 x\,\csc^2 x(1+2\cot^2 x+\cot^4 x) =\cot^4 x\,\csc^2 x(1+\cot^2 x)^2 =\cot^4 x\,\csc^6 x.

Worked Example: Odd Cotangent Exponent

Evaluate \(\int \cot^3 x\,\csc^5 x\,dx\).

The cotangent exponent \(3\) is odd, and a cosecant factor is present. Set \(u=\csc x\), so \(du=-\csc x\cot x\,dx\). After reserving that factor, the remaining \(\cot^2 x\) becomes \(u^2-1\):

\begin{aligned} \int \cot^3 x\,\csc^5 x\,dx &=\int \cot^2 x\,\csc^4 x\, \bigl(\cot x\csc x\,dx\bigr)\\ &=-\int (u^2-1)u^4\,du\\ &=-\int (u^6-u^4)\,du\\ &=-\frac{u^7}{7}+\frac{u^5}{5}+C\\ &=-\frac{\csc^7 x}{7}+\frac{\csc^5 x}{5}+C. \end{aligned}

For the check, differentiate the final expression:

\begin{aligned} \frac{d}{dx}\left(-\frac{\csc^7 x}{7} +\frac{\csc^5 x}{5}\right) &=\csc^6 x(\csc x\cot x) -\csc^4 x(\csc x\cot x)\\ &=\csc^5 x\cot x(\csc^2 x-1)\\ &=\csc^5 x\cot x\cot^2 x\\ &=\cot^3 x\,\csc^5 x. \end{aligned}

Worked Example: Both Substitutions Are Available

Evaluate \(\int \cot^3 x\,\csc^4 x\,dx\). Both the cosecant exponent and the cotangent exponent are even and odd, respectively, so either pattern applies. Use \(u=\cot x\):

\begin{aligned} \int \cot^3 x\,\csc^4 x\,dx &=-\int u^3(1+u^2)\,du\\ &=-\int (u^3+u^5)\,du\\ &=-\frac{u^4}{4}-\frac{u^6}{6}+C\\ &=-\frac{\cot^4 x}{4}-\frac{\cot^6 x}{6}+C. \end{aligned}

Differentiate to verify:

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

The same integrand could be handled with \(u=\csc x\), but both routes are not necessary. Choose one substitution, keep its negative differential sign, and check the resulting antiderivative.

A Common Parity Mistake

A tempting but incorrect move is to use \(u=\csc x\) whenever the cosecant exponent is odd. For example, in \(\int \cot^2 x\,\csc^3 x\,dx\), setting \(u=\csc x\) does not leave a factor \(\cot x\csc x\,dx\): the integrand has \(\cot^2 x\), not a single cotangent factor paired with cosecant. Writing \(du=-\csc x\cot x\,dx\) and cancelling a cotangent factor anyway would introduce a factor that was not present.

This example has an even cotangent exponent and an odd cosecant exponent—the unmatched parity case. Recognizing that neither direct pattern applies is part of choosing correctly. It is better to pause and select another suitable method than to manipulate the factors as if the required derivative were present.

Takeaway: For \(\int \cot^m x\,\csc^n x\,dx\), an even \(n\ge2\) points to \(u=\cot x\); an odd \(m\), with \(n\ge1\), points to \(u=\csc x\). If both patterns apply, choose either. If neither applies, these direct parity substitutions do not provide the polynomial reduction.

Choosing Efficiently

Before expanding powers or integrating, check the exponents and the derivative factors. With an even cosecant exponent, reserve \(\csc^2 x\,dx\) and use \(u=\cot x\). With an odd cotangent exponent, reserve \(\cot x\csc x\,dx\) and use \(u=\csc x\). Then use the appropriate identity to rewrite the remaining factors. This order keeps the substitution transparent and makes sign errors easier to catch.

For positive powers, the two direct patterns overlap whenever \(n\) is even and \(m\) is odd. In such a case, a useful practical choice is the one that leaves the smaller polynomial to expand. Regardless of the choice, the final derivative must reproduce the original integrand on the interval under consideration. In particular, a correct-looking polynomial integral can still lead to a wrong antiderivative if the negative sign in \(du\) is lost.

Check Your Understanding

For each question, identify the applicable substitution pattern before carrying out any integration.

For \(\int \cot^2 x\,\csc^6 x\,dx\), which substitution is suggested by the parity rule, and which differential factor should be reserved?
For \(\int \cot^5 x\,\csc^3 x\,dx\), what substitution converts the remaining cotangent power into a power of the new variable?
For \(\int \cot^3 x\,\csc^2 x\,dx\), do both direct patterns apply, or only one? State the substitution you would choose.
Does \(\int \cot^2 x\,\csc^5 x\,dx\) fit either direct parity pattern? Explain why or why not.
Why does a negative sign appear when using either \(u=\cot x\) or \(u=\csc x\)?

Choosing A Substitution For Mixed Cotangent-Cosecant Powers

What You'll Learn

Why the Exponents Guide the Substitution

The Two Main Parity Patterns

Why These Substitutions Work