Why Reduce a Cosecant Power?
Integrals of high powers of cosecant can be difficult to handle directly. A useful strategy is to express an integral with exponent \(n\) in terms of one with exponent \(n-2\), together with an explicit cotangent term. Repeating that reduction eventually reaches a simple base integral.
The key is to choose an integration-by-parts setup that leaves a factor of \(\csc^2 x\). Its antiderivative is \(-\cot x\), and the identity \(\cot^2 x=\csc^2 x-1\) then connects the resulting integral back to the original power. The negative sign from the antiderivative is essential.
Notation and Conditions
Throughout, \(n\) is an integer with \(n\ge 3\). We work on an interval that contains no integer multiple of \(\pi\). On such an interval, \(\sin x\ne 0\), so \(\csc x\) and \(\cot x\) are defined and differentiable.
Write \(I_n\) for an antiderivative of \(\csc^n x\) on the interval:
This notation describes a family of antiderivatives: any two antiderivatives on the same interval differ by a constant. Accordingly, a reduction formula for \(I_n\) is understood up to an additive constant.
The Reduction Formula
Here \(C\) is an arbitrary constant. The first term is the cotangent term produced by integration by parts. The second term has exponent \(n-2\), so the power decreases by two.
Deriving the Formula by Integration by Parts
Factor the integrand as \(\csc^{n-2}x\csc^2x\). Choose the part with exponent \(n-2\) as \(u\), and integrate \(\csc^2x\) as \(dv\). The derivative identities needed are \(\frac{d}{dx}\csc x=-\csc x\cot x\) and \(\frac{d}{dx}\cot x=-\csc^2x\).
Set \(u=\csc^{n-2}x\) and \(dv=\csc^2x\,dx\). Then \(du=-(n-2)\csc^{n-2}x\cot x\,dx\) and \(v=-\cot x\).
Use \(\int u\,dv=uv-\int v\,du\), keeping the minus sign in \(v=-\cot x\).
Replace \(\cot^2x\) with \(\csc^2x-1\), separating an \(I_n\) term from an \(I_{n-2}\) term.
Move the resulting multiple of \(I_n\) to the left and divide by \(n-1\).
In detail, integration by parts gives
The product inside the second integral is positive before the integration-by-parts subtraction: \((-\cot x)(-(n-2)\csc^{n-2}x\cot x)=(n-2)\csc^{n-2}x\cot^2x\). The subtraction therefore gives the negative sign in the last line. Now use \(\cot^2x=\csc^2x-1\):
Add \((n-2)I_n\) to both sides. This produces \((n-1)I_n\) on the left. Dividing by \(n-1\), which is nonzero for \(n\ge3\), proves the stated formula:
The recurrence’s coefficients and signs follow directly from this collection step. The cotangent term is not optional: it is the boundary term \(uv\) from integration by parts.
Base Cases
The recurrence applies for \(n\ge3\). When reductions reach exponent two or one, the needed antiderivatives are:
The first follows from \(\frac{d}{dx}\cot x=-\csc^2x\). For the second, let \(g(x)=\csc x-\cot x\). Then
Where \(\sin x\ne0\), \(g(x)\ne0\), so \(\frac{d}{dx}\ln|g(x)|=g'(x)/g(x)=\csc x\). These base cases also clarify how the parity of \(n\) affects where a sequence of reductions ends: an even exponent reaches two, while an odd exponent reaches one.
Worked Examples
Worked Example: Apply the Formula When \(n=2\)
The theorem is stated for \(n\ge3\), so \(n=2\) is handled directly as a base case. Integrate using the derivative of cotangent:
Check by differentiating the proposed antiderivative:
Thus the antiderivative differentiates to the original integrand, with the required positive sign.
Worked Example: Use the Reduction for \(n=3\)
Substitute \(n=3\) into the formula. Then \(n-1=2\), \(n-2=1\), and \(\csc^{n-2}x=\csc x\):
Use the base antiderivative for \(\csc x\), absorbing constants into one \(C\):
To check, first differentiate the cotangent product:
Therefore, differentiating the proposed answer gives
The second equality uses \(\cot^2x=\csc^2x-1\), and the final equality follows by cancellation of \(-\frac12\csc x\) with \(+\frac12\csc x\).
Worked Example: Use the Reduction for \(n=4\)
For \(n=4\), \(n-1=3\), \(n-2=2\), and the formula gives
Substitute \(\int\csc^2x\,dx=-\cot x\):
Check by differentiating each term:
The derivative is exactly the original integrand, confirming both the sign and the coefficients.
A Common Sign Error
A frequent mistake is to use \(v=\cot x\) when integrating \(dv=\csc^2x\,dx\). This is incorrect because \(\frac{d}{dx}\cot x=-\csc^2x\), so the correct antiderivative is \(v=-\cot x\). The error can be exposed immediately in the simplest case:
The derivative has the wrong sign. Correcting the antiderivative gives \(-\cot x\), whose derivative is \(+\csc^2x\). In the general derivation, this same sign controls the negative cotangent term and the signs of the integrals after integration by parts.
Why the Reduction Is Useful
Integration by parts transforms the original integral into an equation that contains \(I_n\) on both sides. The identity \(\cot^2x=\csc^2x-1\) makes that possible: one part of the resulting expression reproduces the original power, while the other lowers the exponent by two. Solving the equation isolates the desired antiderivative.
The formula is also a useful check on a proposed result. Its coefficient of the cotangent term must be negative, its denominator must be \(n-1\), and the remaining integral must have exponent \(n-2\) with coefficient \((n-2)/(n-1)\). Those features all come from the integration-by-parts calculation, rather than from a pattern guessed from examples.
Check Your Understanding
Answer these questions using the derivation and conditions above.
- Why is \(v=-\cot x\), rather than \(v=\cot x\), when integrating \(\csc^2x\,dx\)?
- In the reduction formula, by how much does the exponent decrease, and what identity makes that reduction possible?
- Why must the interval of integration avoid integer multiples of \(\pi\)?
- What are the base antiderivatives for \(\csc x\) and \(\csc^2x\)?
- Differentiate the cotangent term in the formula for \(n=3\) and identify the sign that it contributes.
