Why Pair an Odd Cotangent Power With Cosecant?
Some trigonometric integrals become manageable when one factor is saved to match the derivative of a substitution. Here, an odd power of cotangent contains one factor of \(\cot x\) after the other cotangent factors are paired. If the integrand also contains at least one factor of \(\csc x\), those saved factors combine as \(\csc x\cot x\,dx\), which is closely related to the derivative of \(\csc x\).
The remaining cotangent factors occur in an even power. The identity \(\cot^2 x=\csc^2 x-1\) converts those factors into an expression in cosecant alone. The substitution then turns the integral into a polynomial integral.
The restriction \(n\ge 1\) ensures that the integrand contains a cosecant factor to pair with the saved cotangent. The exponent \(2k+1\) is odd, so after saving one \(\cot x\), the remaining cotangent power is even. These features—not merely the presence of cotangent and cosecant—are what make this particular substitution pattern work directly.
Set Up the Substitution
First separate one \(\csc x\cot x\) factor. Then convert each pair of remaining cotangent factors using the identity \(\cot^2 x=\csc^2 x-1\). The cosecant substitution is \(u=\csc x\), with differential \(du=-\csc x\cot x\,dx\). The minus sign must be retained.
Because \(k\) is a nonnegative integer, \((u^2-1)^k\) is a polynomial. Multiplication by \(u^{n-1}\) still gives a polynomial, so each term can be integrated by the power rule. This is the central advantage of choosing the substitution after reserving the matched factor.
A General Formula and Its Justification
The binomial theorem expands the remaining power as \((u^2-1)^k=\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}u^{2j}\). Thus the integrand in \(u\) is a finite sum of powers. Since \(n\ge1\), every exponent \(n-1+2j\) is nonnegative, and the corresponding antiderivative denominator \(n+2j\) is nonzero.
Here \(C\) is an arbitrary constant. To justify the formula, start with the substitution expression and expand its polynomial. Integrating each term gives \(-\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}u^{n+2j}/(n+2j)+C\). Substituting \(u=\csc x\) gives the displayed result. This derivation uses only the binomial theorem, the power rule, the identity for \(\cot^2 x\), and the derivative of cosecant.
The formula can also be checked directly by differentiation. The derivative of \(\csc^m x\), for a positive integer \(m\), is \(-m\csc^m x\cot x\). Differentiating the formula therefore gives \(\cot x\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}\csc^{n+2j}x\). The sum is \(\csc^n x(\csc^2 x-1)^k=\csc^n x\cot^{2k}x\). Multiplying by the remaining \(\cot x\) recovers \(\cot^{2k+1}x\csc^n x\), as required.
Worked Examples
Worked Example: Integrate \(\cot^3 x\,\csc^2 x\)
Here \(2k+1=3\), so \(k=1\), and \(n=2\). Save one factor \(\csc x\cot x\,dx\); the remaining cotangent factor is \(\cot^2 x\).
To check, differentiate the result. The derivative of \(-\csc^4 x/4\) is \(\csc^4 x\cot x\); the derivative of \(\csc^2 x/2\) is \(-\csc^2 x\cot x\). Their sum is \(\csc^2 x\cot x(\csc^2 x-1)=\csc^2 x\cot x\cot^2 x=\cot^3 x\csc^2 x\). The derivative matches the integrand.
Worked Example: Integrate \(\cot^5 x\,\csc^3 x\)
Now \(2k+1=5\), so \(k=2\), and \(n=3\). Reserve \(\csc x\cot x\,dx\); rewrite the other four cotangent factors as \((\cot^2 x)^2\).
For the check, differentiate term by term: \[ \frac{d}{dx}\left(-\frac{\csc^7 x}{7}\right)=\csc^7 x\cot x,\quad \frac{d}{dx}\left(\frac{2\csc^5 x}{5}\right)=-2\csc^5 x\cot x,\quad \frac{d}{dx}\left(-\frac{\csc^3 x}{3}\right)=\csc^3 x\cot x. \] Their sum is \(\csc^3 x\cot x(\csc^4 x-2\csc^2 x+1)=\csc^3 x\cot x(\csc^2 x-1)^2=\csc^3 x\cot x\cot^4 x\), which is the original integrand.
Worked Example: Integrate \(\cot^3 x\,\csc x\)
This case has \(k=1\) and \(n=1\), so the substitution still leaves a polynomial. The saved factor is \(\csc x\cot x\,dx\), while the remaining cotangent square becomes \(u^2-1\).
Differentiate to check: \[ \frac{d}{dx}\left(-\frac{\csc^3 x}{3}+\csc x\right) =\csc^3 x\cot x-\csc x\cot x =\csc x\cot x(\csc^2 x-1) =\csc x\cot^3 x. \] This is exactly the given integrand.
The Sign and the Conditions Matter
A frequent error is to lose the minus sign in \(du=-\csc x\cot x\,dx\). For example, in the first worked example, incorrectly treating \(\csc x\cot x\,dx\) as \(du\) leads to \(\int (u^2-1)u\,du=u^4/4-u^2/2+C\). Differentiating that proposed answer with respect to \(x\) gives \(-\csc^4 x\cot x+\csc^2 x\cot x=-\cot^3 x\csc^2 x\), the negative of the required integrand. The correction is to use \(\csc x\cot x\,dx=-du\), which reverses the sign of the entire \(u\)-integral.
Another important check is whether the factors really match. If there is no cosecant factor available, one cannot simply claim that a factor \(\csc x\cot x\,dx\) has been reserved. Likewise, if the cotangent exponent is even, removing one cotangent would leave an odd power rather than the even power needed for direct rewriting with \(\cot^2 x=\csc^2 x-1\). Other strategies may apply to other integrals, but this formula is specifically for an odd cotangent power paired with a positive cosecant power.
All expressions involving \(\csc x\) or \(\cot x\) require \(\sin x\ne0\). An antiderivative obtained by this method is understood on an interval that does not cross a point where \(\sin x=0\). The constant of integration may differ on separate such intervals.
How to Apply the Method Reliably
Confirm that the cotangent exponent is \(2k+1\) for a nonnegative integer \(k\), and that the cosecant exponent \(n\) is at least one.
Set aside one factor \(\csc x\cot x\,dx\), and explicitly record \(du=-\csc x\cot x\,dx\) for \(u=\csc x\).
Use \(\cot^{2k}x=(\cot^2 x)^k=(\csc^2 x-1)^k\).
Substitute \(u=\csc x\), expand the polynomial, integrate each power, and differentiate the final expression to check the result.
This approach is useful because it turns a trigonometric integral into a finite algebraic calculation while keeping the substitution and the identity visibly connected. Writing the factorization before substituting helps prevent both a missing factor and a sign error. The general formula packages the same steps efficiently, but for a particular integral it is often clearer to show the substitution and polynomial expansion directly.
Check Your Understanding
Use the factorization and the substitution carefully. In each case, check the proposed antiderivative by differentiation.
- Evaluate \(\int \cot x\,\csc^2 x\,dx\).
- For \(\int \cot^3 x\,\csc^4 x\,dx\), identify \(k\), \(n\), and the polynomial in \(u\) before integrating.
- Explain why the differential \(\csc x\cot x\,dx\) contributes a minus sign when \(u=\csc x\).
- State the conditions on \(k\), \(n\), and \(x\) used in the general formula.
- Differentiate \(-\csc^3 x/3+\csc x\) and simplify the result using \(\csc^2 x-1=\cot^2 x\).
