Powers of Multiple-Angle Sines and Cosines

Why the Multiple Angle Matters

A power such as \(\sin^5(3x)\) has two features that affect the calculation: the exponent and the factor multiplying \(x\) inside the trigonometric function. The exponent determines whether an odd-power substitution or an even-power identity is convenient. The factor inside the angle changes the scale of the integral and, for a definite integral, changes the limits.

The first step is therefore to rescale the angle. After that, use the familiar parity-based methods for sine and cosine powers. The extra care is to carry the scale factor through every step rather than treating the multiple-angle expression as if its angle were simply \(x\).

Rescale the Angle

For \(k\ne 0\), set \(u=kx\). Then \(du=k\,dx\), so \(dx=du/k\). Thus, for an indefinite integral, the scaling rule is

The factor \(1/k\) is essential. When \(k\) is negative, it remains negative; it should not be replaced by \(1/|k|\).

For a definite integral, the same substitution changes the endpoints as well as the differential. If \(x\) runs from \(a\) to \(b\), then \(u\) runs from \(ka\) to \(kb\):

The corresponding cosine formula is identical with \(\sin^m u\) replaced by \(\cos^m u\). Keep the transformed endpoints in that order, including when \(k<0\).

Odd Powers: Save One Factor

When the exponent is odd, write it as \(m=2r+1\). For a sine power, keep one factor of \(\sin u\) and rewrite the rest using \(\sin^2u=1-\cos^2u\). This leaves a polynomial in \(\cos u\), and the remaining factor \(\sin u\,du\) is suited to the substitution \(v=\cos u\).

Here is the general sine calculation. Expanding the polynomial gives a formula for any nonnegative integer \(r\):

The minus sign appears because \(dv=-\sin u\,du\). For an odd cosine power, the same expansion uses \(v=\sin u\) and has a positive sign in front of the resulting sum.

Even Powers: Reduce the Power

If the exponent is even, there is no single sine or cosine factor left over to pair directly with the derivative of the other function. Instead, use the power-reduction identities. The basic identities are

For example, squaring either identity and using the double-angle identity for \(\cos^2(2u)\) gives a sum of a constant and cosines of multiple angles. The resulting terms can be integrated one at a time. Higher even powers can be reduced by applying the same identities repeatedly.

Why the Scaling Rule Works

The change of variable is not an extra trick added to the power methods. It follows from the chain rule. If \(G'(u)=g(u)\), then

Thus an antiderivative of \(g(u)\), evaluated at \(u=kx\), must be multiplied by \(1/k\). For a definite integral, the change-of-variable rule also changes the endpoints: \(x=a\) corresponds to \(u=ka\), and \(x=b\) corresponds to \(u=kb\). This explains both the scaling factor and the transformed limits.

For example, if \(F'(u)=\sin^3u\), then differentiating \(F(kx)/k\) gives \(\sin^3(kx)\). The definite integral from \(a\) to \(b\) is \(F(kb)/k-F(ka)/k\), which is exactly \((1/k)\int_{ka}^{kb}\sin^3u\,du\).

Definite Integrals: Transform Both Limits

For an odd power, the substitution can be carried out directly in the definite integral. The limits for \(v=\cos u\) or \(v=\sin u\) must be transformed too. The next examples show two common cases: an odd power handled by substitution and an even power handled by power reduction.

A Common Scaling Error

A frequent mistake is to substitute \(u=kx\) but omit the factor \(1/k\). For instance, the expression

is not an antiderivative of \(\sin^3(2x)\). Differentiating it gives \(2\sin^3(2x)\), not \(\sin^3(2x)\), because the chain rule contributes a factor of \(2\). The correction is to divide the expression by \(2\):

Differentiating this corrected expression gives \(\sin(2x)-\cos^2(2x)\sin(2x)=\sin^3(2x)\). For a definite integral, omitting the transformed endpoints is a related error: both endpoints must be multiplied by \(k\), not merely the integrand adjusted.

A Practical Decision Process

The method matters because a multiple-angle power can otherwise look more complicated than it is. Rescaling separates the frequency from the power calculation. Once the scale factor and limits are handled correctly, the parity methods apply as usual.