Closed Products from Trigonometric Power Recurrences

From a recurrence to a finite expression

Reduction formulas replace an antiderivative with one involving a lower power. Used repeatedly, they turn an apparently open-ended calculation into a finite sum: each step lowers the exponent by two, until a base case is reached. The coefficients accumulated along the way form products of simple ratios.

This tutorial develops that process for nonnegative integer powers of sine and cosine. The parity of the starting exponent determines whether the last integral is a zeroth or first power. The formulas below package both the recurring boundary terms and the final base-case contribution.

The recurrences and their base cases

For \(n\ge 2\), the sine- and cosine-power reduction formulas are

They can be used repeatedly because each application reduces the exponent by two. The process ends at one of four base integrals:

These values follow by integrating \(1\), \(\sin x\), and \(\cos x\). When applying a recurrence, it is helpful to distinguish two kinds of terms: a boundary term produced at the current exponent, and a lower-power antiderivative multiplied by a new ratio. That ratio must be carried forward at every step.

How the product coefficients arise

After \(j\) reductions from exponent \(n\), the coefficient multiplying the remaining antiderivative is the product of the \(j\) ratios encountered so far. Define

The product with no factors, \(P_{n,0}\), is defined to equal \(1\). Thus \(P_{n,j+1}=P_{n,j}\frac{n-2j-1}{n-2j}\). This relation explains why the next reduction multiplies the accumulated coefficient by exactly the next recurrence ratio.

For instance, \(P_{6,1}=5/6\), \(P_{6,2}=(5/6)(3/4)=5/8\), and \(P_{6,3}=(5/8)(1/2)=5/16\). The last product is the coefficient of \(I_0\) in the fully reduced expression for \(I_6\).

Closed forms for even and odd powers

When the exponent is even, the recurrence ends at exponent zero. When it is odd, it ends at exponent one. Writing the accumulated ratios as \(P_{n,j}\) gives the following finite formulas.

For cosine powers, the corresponding formulas are

When \(m=0\), a sum whose upper limit is \(-1\) is empty and equals zero. This convention makes the odd formulas include the base cases \(I_1=-\cos x+C\) and \(J_1=\sin x+C\), while the even formulas include \(I_0=J_0=x+C\).

Why the closed forms follow

Consider the sine recurrence. At exponent \(n\), its boundary term is \(-\sin^{n-1}x\cos x/n\), and the remaining antiderivative has coefficient \((n-1)/n\). After \(j\) reductions, the current exponent is \(n-2j\), and the remaining antiderivative has coefficient \(P_{n,j}\). Its next boundary term therefore has coefficient \(P_{n,j}\) and denominator \(n-2j\), as in the finite sums above.

This argument proves the sine formulas: every reduction contributes exactly the indicated summand and updates the coefficient by the stated product ratio. The cosine formulas follow by the same coefficient updates, using the cosine recurrence and its boundary term instead. In each case the exponent falls by two, so the process terminates after finitely many steps.

Worked examples

A common coefficient error

A frequent mistake is to add the next boundary term but forget to multiply the remaining integral by the recurrence ratio. For example, the first step for \(I_6\) is sometimes written incorrectly as

This omits the factor \(5/6\). The correct recurrence is

Consequently, every term obtained by expanding \(I_4\)—including its \(x\)-term—must also be multiplied by \(5/6\). The product notation is useful precisely because it records these accumulated factors and makes omissions easier to detect.

Using and checking the formulas

These expressions are useful when a problem asks for an antiderivative in terms of powers of sine or cosine and a finite collection of lower powers. They also make the structure of repeated reduction visible: the boundary terms form a finite sum, and the base case contributes either a multiple of \(x\) or a final first-power term.

They are not a reason to skip verification. At each step, check that the exponent decreases by two, that the denominator matches the current exponent, and that the remaining integral receives the correct ratio. After expanding, differentiate the result and simplify using \(\sin^2x+\cos^2x=1\). That check confirms both the signs and the accumulated coefficients.