Reducing Fourth and Higher Even Powers

Why Reduce an Even Power?

When a trigonometric integrand contains an even power such as \(\sin^4 x\) or \(\cos^6 x\), the usual substitution used for an odd power is not directly available: there is no single factor of \(\sin x\) or \(\cos x\) left over after rewriting the rest using \(\sin^2 x+\cos^2 x=1\). A useful alternative is to replace the even power by a sum of a constant and cosines of multiple angles.

Once the expansion is in that form, integration is direct. The main care is to keep the frequency of each cosine: integrating \(\cos(2jx)\) introduces a factor \(1/(2j)\). For powers of sine, the multiple-angle terms also alternate in sign.

The Fourth-Power Identities

The basic double-angle identity is \(\cos(2x)=2\cos^2 x-1\). Rearranging gives \(\cos^2 x=(1+\cos(2x))/2\). Likewise, \(\sin^2 x=(1-\cos(2x))/2\). Squaring these expressions and reducing the resulting squared cosine gives the fourth-power formulas.

For cosine, first square its power-reduction identity:

Apply the same identity to \(\cos^2(2x)\), so that \(\cos^2(2x)=(1+\cos(4x))/2\). Substitution and collection of terms give:

For sine, square \(\sin^2 x=(1-\cos(2x))/2\) and use the same reduction for \(\cos^2(2x)\):

The constant terms agree, and the coefficient of \(\cos(4x)\) agrees. The \(\cos(2x)\) terms have opposite signs. This sign difference comes from the minus sign in the formula for \(\sin^2 x\).

A Formula for Every Positive Even Power

The fourth-power identities extend to every positive integer power \(n\). The cosine expansion is:

Here \(\binom{2n}{k}\) is the binomial coefficient. The cosine formula has only positive coefficients. In the sine formula, the coefficient of \(\cos(2jx)\) has sign \((-1)^j\): negative for odd \(j\), positive for even \(j\).

Why the General Expansion Works

A compact derivation uses Euler’s identity \(e^{ix}=\cos x+i\sin x\), which gives \(\cos x=(e^{ix}+e^{-ix})/2\). Apply the binomial theorem to the \(2n\)-th power:

The middle term, where \(r=n\), has frequency zero and equals \(\binom{2n}{n}/2^{2n}\). Every other term pairs with the term whose index is \(2n-r\). If \(r=n-j\), the paired terms have equal binomial coefficients and opposite frequencies:

Adding the constant term and all \(n\) pairs proves the cosine formula. To obtain the sine formula, use \(\sin x=\cos(x-\pi/2)\). In each cosine term this shift gives \(\cos(2j(x-\pi/2))=(-1)^j\cos(2jx)\), while the constant term is unchanged. That proves the stated sine expansion as well.

This derivation justifies the general coefficients and signs. In calculation, the same expansions can be built up from power-reduction and double-angle identities, as in the fourth-power derivations. For the sixth power, the formula gives:

Integrating the General Expansion

The general formula also gives a direct antiderivative. The constant term integrates to a multiple of \(x\), and each \(\cos(2jx)\) integrates to \(\sin(2jx)/(2j)\). Thus, for positive integer \(n\):

These formulas are not separate integration tricks. They apply the same term-by-term integration rule to the corresponding power expansions. They are especially useful when the exponent is large and writing each coefficient by hand would invite arithmetic errors.

A Common Frequency Error

A frequent mistake is to integrate \(\cos(4x)\) as \(\sin(4x)\), forgetting that the inside angle has derivative \(4\). For instance, this incorrect step loses a factor:

The correction is to divide by \(4\):

Differentiating the incorrect expression would give \(\frac{1}{2}\cos(4x)\), not \(\frac{1}{8}\cos(4x)\). Differentiating the corrected expression gives \(\frac{1}{8}\cos(4x)\), as required. Always check the frequency when integrating a multiple-angle term.

Using the Method Reliably

For an even power of sine or cosine, first identify the exponent as \(2n\). Expand into a constant and cosine terms with frequencies \(2x,4x,\ldots,2nx\). For a sine power, check the alternating signs; for a cosine power, the coefficients are positive. Then integrate term by term, dividing each sine coefficient by the corresponding frequency. Finally, differentiate the proposed antiderivative to verify it.

The reduction is useful because it changes a power into a short sum of familiar functions without changing the integration problem into a more complicated one. It also makes clear why an even-power antiderivative contains a term proportional to \(x\): the expansion always has a nonzero constant term.