Even Cosecant Powers By Cotangent Substitution

The Cotangent Analogue of the Secant Method

For integrals of the form \(\int\cot^mx\,\csc^{2k}x\,dx\), the cotangent plays the role that the tangent played with even powers of secant. The key fact is \(\frac{d}{dx}\cot x=-\csc^2x\), so \(\csc^2x\,dx=-du\) when \(u=\cot x\). Since the integrand has an even power of cosecant, one factor \(\csc^2x\) can be reserved for \(du\), and the remaining \(\csc^{2k-2}x=(\csc^2x)^{k-1}\) can be written as a polynomial in \(\cot x\) with the identity \(\csc^2x=1+\cot^2x\).

The only new feature compared with the tangent case is the minus sign in the derivative of the cotangent. It must be tracked in every step.

Deriving the Substitution

Write \(\csc^{2k}x=\csc^2x\cdot(\csc^2x)^{k-1}=\csc^2x\,(1+\cot^2x)^{k-1}\). With \(u=\cot x\) and \(du=-\csc^2x\,dx\),

The integrand on the right is a polynomial, so expand and integrate term by term. For \(k=1\) the factor \((1+u^2)^{0}\) is just \(1\) and the integral is \(-\frac{u^{m+1}}{m+1}\).

Checking by Differentiation

To check \(-\frac{\cot^{m+1}x}{m+1}\), differentiate: the power rule gives \(-\cot^mx\), and the chain rule multiplies by \(\frac{d}{dx}\cot x=-\csc^2x\). The two minus signs cancel and the result is \(\cot^mx\csc^2x\). For higher \(k\), differentiate each polynomial term the same way.

Worked Examples

A Reliable Procedure

Common Mistakes

The first mistake is the lost sign: \(\int\csc^2x\,dx=-\cot x+C\), not \(+\cot x\). The second is to use \(\csc^2x=1-\cot^2x\) or to confuse the identity with \(\sec^2x=1+\tan^2x\); the correct cosecant identity is \(\csc^2x=1+\cot^2x\). The third is to reserve too many factors, such as \(\csc^4x\), which cannot be written as a derivative of a single function of \(\cot x\).

Finally, for definite integrals note that \(\cot x\) is undefined at \(x=0\) and \(x=\pi\). The interval of integration must lie within one interval \((n\pi,(n+1)\pi)\), or the integral is improper.

Quick Reference

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Find an antiderivative of \(\cot^2x\csc^4x\). Hint: remember the overall minus sign. Answer: \(-\frac{\cot^3x}{3}-\frac{\cot^5x}{5}\)
2. Find an antiderivative of \(\cot x\csc^4x\). Hint: use \(u=\cot x\) and expand \(u(1+u^2)\). Answer: \(-\frac{\cot^2x}{2}-\frac{\cot^4x}{4}\)
3. Evaluate \(\int_{\pi/4}^{\pi/2}\cot^2x\csc^4x\,dx\). Hint: \(\cot\frac\pi2=0\) and \(\cot\frac\pi4=1\). Answer: \(\frac{8}{15}\)
4. Evaluate \(\int_{\pi/4}^{\pi/2}\csc^4x\,dx\). Hint: use \(-\cot x-\frac{\cot^3x}{3}\). Answer: \(\frac43\)

Comparison with the Secant Case

The two families are mirror images. The sign difference comes entirely from the fact that cotangent is a decreasing function whose derivative is negative, while tangent is increasing. Knowing one family gives the other by changing the function names and attaching minus signs.

Interpreting the Result

On \((0,\pi)\) the integrand \(\cot^mx\csc^{2k}x\) is positive for even \(m\) and has the sign of \(\cot x\) for odd \(m\). The antiderivative is a decreasing function of \(\cot x\), which itself decreases from \(+\infty\) to \(-\infty\) across the interval, so the antiderivative as a function of \(x\) increases when the integrand is positive. Near \(x=0\) and \(x=\pi\) the integrals grow without bound, because \(\csc x\) becomes infinite. This is the reason the endpoints \(0\) and \(\pi\) must be avoided in definite integrals.

A Numerical Spot Check

Test \(\int_{\pi/4}^{\pi/2}\csc^4x\,dx=\frac43\) with the trapezoid rule on one interval. The heights are \(\csc^4\frac\pi4=4\) and \(\csc^4\frac\pi2=1\). The estimate is \(\frac{\pi}{4}\cdot\frac{4+1}{2}\approx1.963\), which is higher than \(\frac43\approx1.333\) because the integrand is convex and decreasing. Using two intervals, with midpoint height \(\csc^4\frac{3\pi}{8}\approx1.3726\), the estimate drops to \(\frac{\pi}{8}\bigl(\frac{4+1.373}{2}+\frac{1.373+1}{2}\bigr)\approx1.521\), moving toward the exact value as expected.

What Happens with an Odd Power of Cosecant

The method above requires an even power of cosecant. If the cosecant power is odd and the cotangent power is odd, a different pairing is used: reserve \(\csc x\cot x\,dx=-d(\csc x)\) and set \(u=\csc x\). For example, \(\int\cot^3x\csc^3x\,dx\) is solved by writing \(\cot^2x=\csc^2x-1\) and substituting \(u=\csc x\). That case belongs to a later tutorial. For now, the rule is simple: look at the power of cosecant first. An even power signals the cotangent substitution, and an odd power with an odd cotangent signals the cosecant substitution.

Keeping this decision rule in mind prevents the common error of forcing \(u=\cot x\) onto an integrand where it produces a leftover factor that is not a polynomial in \(u\).

Where This Fits in the Course

This tutorial belongs to the module “Tangent and Secant Powers”. The tutorial just before it is “Derivative Identities Behind Cotangent And Cosecant Integrals,” and the one that follows is “Odd Cotangent Powers Paired With Cosecant Factors.” It mirrors the secant method and completes the family of even-power trigonometric substitutions, so the techniques here carry over almost unchanged once the signs are tracked carefully.