Integrating Mixed Even Powers

Why Mixed Even Powers Need a Different Approach

An integral such as \(\int \sin^4 x\cos^2 x\,dx\) has no single odd power that can be separated to pair with the derivative of the other trigonometric function. A useful approach is to rewrite one factor using \(\sin^2 x+\cos^2 x=1\), then reduce the resulting even powers. Alternatively, double-angle identities can turn the product into a polynomial in \(\cos 2x\).

Both methods ultimately use familiar identities, but their expansions can look quite different. The best choice depends on the exponents and on which intermediate expressions are easiest to manage.

Method 1: Rewrite One Factor and Reduce

Since \(\sin^2x=1-\cos^2x\), the integrand can be written as \[ \sin^{2m}x\cos^{2n}x =(1-\cos^2x)^m\cos^{2n}x. \] Expanding the power with the binomial theorem gives a sum of even powers of cosine:

Thus the original integral becomes a finite sum of integrals of even powers of cosine. One can instead use \(\cos^2x=1-\sin^2x\) and reduce to even powers of sine. Choose the version that produces fewer terms or lower powers.

For clarity, define \(C_k(x)\) to be an antiderivative of \(\cos^{2k}x\). For \(k=0\), \(C_0(x)=x\). For \(k\geq 1\), the even-power reduction formula is:

This formula allows repeated reduction until the integral reaches \(C_0(x)=x\). It follows by integration by parts: take \(u=\cos^{2k-1}x\) and \(dv=\cos x\,dx\). Then \(v=\sin x\) and \(du=-(2k-1)\sin x\cos^{2k-2}x\,dx\), so

Collecting the \(C_k\) terms gives \(2kC_k=\sin x\cos^{2k-1}x+(2k-1)C_{k-1}\), which is the stated recurrence. The same reasoning applies to even powers of sine, or those can be converted using \(\sin^2x=1-\cos^2x\).

Worked Example: Two Expansions for \(\sin^2x\cos^2x\)

Method 2: Convert the Product Using Double Angles

The identities \(\sin^2x=(1-\cos 2x)/2\) and \(\cos^2x=(1+\cos 2x)/2\) give a direct alternative:

This is a polynomial in \(\cos 2x\). Expand it, then integrate its terms. For an even power \(\cos^{2k}(2x)\), use the even-power reduction formula with angle \(2x\), accounting for the factor from differentiating that angle. If an odd power occurs, save one cosine factor and substitute \(u=\sin 2x\). These are ways to evaluate the polynomial terms; no product-to-sum identity is required.

Why does this method work? Each original factor is a power of a double-angle expression, so multiplying the expressions produces an ordinary finite polynomial. Integration is linear, meaning the integral of the polynomial is the sum of the integrals of its terms. The identities hold pointwise, so the polynomial and the original trigonometric product have exactly the same values for every \(x\).

Worked Example: Reduce to Even Powers of Cosine

How to Choose and Compare the Methods

Factor reduction is often efficient when rewriting one factor creates only a few even-power terms. In the \(\sin^4x\cos^2x\) example, the result was a sum of three cosine powers. For larger exponents, however, the binomial expansion can produce many terms, with alternating signs.

The double-angle approach can make the integrand compact, especially when the product contains repeated pairs such as \(\sin^2x\cos^2x\). But raising the double-angle expressions to higher powers may also create a large polynomial. Neither method is always shorter. A useful first step is to inspect both the exponents and any simple identities already present in the product.

The two general expansions are consistent because they start with identities for the same factors. Rewriting \(\sin^2x\) as \(1-\cos^2x\) and rewriting both squared functions with double angles do not change the integrand; they only arrange its terms differently. When two resulting antiderivatives are correct, their difference is constant on any interval where they are compared.

A Common Factor-of-Two Error

A frequent mistake is to combine the two double-angle identities without multiplying their denominators. For example, the correct calculation is

If the denominator \(4\) is mistakenly written as \(2\), the result becomes \((1-\cos 4x)/4\), which is twice the true integrand. Integrating that incorrect expression gives \(x/4-\sin 4x/16\); its derivative is \((1-\cos 4x)/4\), not \(\sin^2x\cos^2x\). Differentiation exposes the error immediately.