Choosing a Substitution by Parity
Mixed cotangent–cosecant integrals are easiest when the factors are grouped to match the derivative of one trigonometric function. The relevant derivatives are \( \frac{d}{dx}(\cot x)=-\csc^2 x \) and \( \frac{d}{dx}(\csc x)=-\csc x\cot x \). Each contains a minus sign. Recognizing which factor can be set aside for one of these derivatives is the key to choosing a substitution.
For the integrals in this lesson, parity gives a practical guide. An even power of cosecant supplies a factor of \( \csc^2 x\,dx \), which pairs with \(u=\cot x\). When the cotangent power is odd and the cosecant power is odd, a factor of \( \cot x\csc x\,dx \) pairs with \(u=\csc x\). In either case, the remaining factors can be rewritten as a polynomial in \(u\).
The Two Substitution Patterns
Suppose first that an integrand has an even cosecant power. Separate one factor of \( \csc^2 x \), and set \(u=\cot x\). Then \(du=-\csc^2 x\,dx\), while every remaining factor of \( \csc^2 x \) can be replaced by \(1+u^2\). The integral becomes a polynomial integral in \(u\).
If instead both powers are odd, separate one factor of \( \cot x\csc x \), and set \(u=\csc x\). Then \(du=-\cot x\csc x\,dx\). The remaining cotangent power is even, so replace each \( \cot^2 x \) by \(u^2-1\). The remaining cosecant factors are already powers of \(u\).
| Pattern to look for | Substitution | Derivative factor to set aside |
|---|---|---|
| Even power of \( \csc x \) | \(u=\cot x\) | \(\csc^2 x\,dx=-du\) |
| Odd powers of both \( \cot x \) and \( \csc x \) | \(u=\csc x\) | \(\cot x\csc x\,dx=-du\) |
These are selection rules for the patterns considered here, not a claim that every mixed cotangent–cosecant integral fits one of them. In each example, the chosen substitution must account for a complete derivative factor, and all remaining factors must be expressible using the new variable.
Why the Substitutions Work
The substitution is a change of variable, not a guess that changes the value of the integral. For example, if \(u=\cot x\), then \(du=-\csc^2 x\,dx\). A factor \( \csc^2 x\,dx \) in the original integral can therefore be replaced by \(-du\), and \( \cot x \) becomes \(u\). The identity \( \csc^2 x=1+\cot^2 x \) converts any remaining cosecant-square factors into \(1+u^2\). The resulting integral is an ordinary polynomial antiderivative, which can then be expressed back in terms of \(x\).
The same reasoning applies to \(u=\csc x\): differentiation gives \(du=-\csc x\cot x\,dx\), and the identity \( \cot^2 x=\csc^2 x-1 \) converts the remaining even cotangent power into a polynomial in \(u\). Thus the identities and the derivative factors together explain why parity guides the method.
Look for an even cosecant power, or for odd powers of both cotangent and cosecant.
Use \( \csc^2 x\,dx \) for \(u=\cot x\), or \( \cot x\csc x\,dx \) for \(u=\csc x\).
Apply \( \csc^2 x=1+\cot^2 x \) or \( \cot^2 x=\csc^2 x-1 \), as appropriate.
Integrate the polynomial, substitute back, and check by differentiating.
Worked Examples
Worked Example: An Even Cosecant Power
Evaluate \( \int \cot^2 x\csc^4 x\,dx \). The cosecant exponent is even. Separate one \( \csc^2 x \) factor for the derivative of cotangent:
Set \(u=\cot x\), so \(du=-\csc^2 x\,dx\). The other cosecant-square factor becomes \( \csc^2 x=1+\cot^2 x=1+u^2 \). Therefore,
Check by differentiating. Since \( \frac{d}{dx}(\cot x)=-\csc^2 x \),
The final equality uses \(1+\cot^2 x=\csc^2 x\), so the derivative matches the integrand.
Worked Example: Odd Cotangent and Cosecant Powers
Evaluate \( \int \cot^3 x\csc^3 x\,dx \). Both exponents are odd. Separate one \( \cot x\csc x\,dx \) factor, which is the negative differential for \(u=\csc x\):
Set \(u=\csc x\). Then \(du=-\cot x\csc x\,dx\), and \( \cot^2 x=\csc^2 x-1=u^2-1 \). Thus
For the check, differentiate the answer and use \(u'=-\csc x\cot x\):
Here \( \csc^2 x-1=\cot^2 x \), so the derivative is the original integrand.
Worked Example: A Single Cosecant-Square Differential
Evaluate \( \int \cot^4 x\csc^2 x\,dx \). The factor \( \csc^2 x\,dx \) is already present, so set \(u=\cot x\) directly:
Differentiating verifies the result:
No identity conversion is needed in this example because all factors other than the saved differential are already powers of \( \cot x \).
A Common Sign Error
Both substitutions have a negative differential: \(du=-\csc^2 x\,dx\) when \(u=\cot x\), and \(du=-\cot x\csc x\,dx\) when \(u=\csc x\). Omitting that minus sign changes the antiderivative. For instance, in the third example, the incorrect calculation
has the wrong sign. Differentiating the proposed answer exposes the error:
The correction is to carry the negative differential through the integral: \( \int \cot^4 x\csc^2 x\,dx=-\int u^4\,du \). The derivative check is especially useful here because it catches a sign error immediately.
Domains and the Constant of Integration
Cotangent and cosecant are undefined wherever \( \sin x=0 \), that is, at integer multiples of \( \pi \). Each antiderivative above is therefore understood on an interval that does not contain such a point. On any such interval the substitutions and derivative checks are valid. The symbol \(C\) denotes an arbitrary constant on the interval.
When applying this method, the central task is to match a factor in the integrand with the derivative of the proposed substitution. Parity helps reveal that factor; it does not remove the need to check the algebra or the sign. Rewriting the leftover powers carefully and differentiating the final expression confirms that the result is correct.
Check Your Understanding
For each question, identify the substitution pattern before carrying out any integration.
- For \( \int \cot^2 x\csc^6 x\,dx \), which differential factor would you save, and which substitution would you use?
- For \( \int \cot^5 x\csc^3 x\,dx \), what factor pairs with the derivative of \( \csc x \)?
- Rewrite \( \cot^2 x \) in terms of \(u=\csc x\).
- Why does a substitution using \(u=\cot x\) introduce a negative sign?
- What should you check after finding a mixed cotangent–cosecant antiderivative?
