Integrating Products with Both Exponents Odd

The pattern when both powers are odd

Integrals of products of powers of sine and cosine become especially direct when both exponents are odd. The reason is that an odd power contains one factor that can be saved for substitution; after removing that factor, the remaining power is even and can be rewritten using \(\sin^2 x+\cos^2 x=1\).

For example, in \(\sin^5 x\cos^3 x\), we can save one factor of cosine and rewrite the other four cosine factors as powers of \(1-\sin^2 x\). Then \(u=\sin x\) turns the integral into a polynomial integral. Alternatively, we can save one sine factor and substitute \(u=\cos x\). Both choices work; the sign from \(du=-\sin x\,dx\) must be handled carefully.

The two substitution choices

Write the odd exponents as \(m=2r+1\) and \(n=2s+1\), where \(r\) and \(s\) are nonnegative integers. To use \(u=\cos x\), save one sine factor. The other \(2r\) sine factors form an even power, so they can be replaced by \(1-\cos^2 x\).

The minus sign in \(du\) means that the resulting integral has an overall negative sign. The other choice saves one cosine factor and uses \(u=\sin x\):

There is no extra minus sign in this version. In either case, substitution leaves an ordinary polynomial in \(u\), which can be integrated term by term.

General formula and why it works

The binomial theorem expands the even power after substitution. Using \(u=\cos x\) gives one formula; using \(u=\sin x\) gives another. These formulas are not separate tricks: they record the same substitution procedure for arbitrary positive odd exponents.

Here \(C\) is an arbitrary constant of integration. The two displayed expressions can look different, but each is an antiderivative of the same function, so they can differ only by a constant on the real line.

To justify the first formula, save \(\sin x\) and substitute \(u=\cos x\). The binomial theorem gives \((1-u^2)^r=\sum_{k=0}^{r}(-1)^k\binom{r}{k}u^{2k}\). Multiplying by \(u^{2s+1}\), integrating each power, and using \(du=-\sin x\,dx\) produces the stated negative sum. For the second formula, save \(\cos x\), substitute \(u=\sin x\), and expand \((1-u^2)^s\); the differential has no minus sign, giving the stated positive sum. This proves both formulas and explains their signs.

For a shorter calculation, compare the number of terms. The cosine substitution expands a power indexed by \(r\), producing \(r+1\) terms; the sine substitution produces \(s+1\) terms. Choosing the smaller of \(r\) and \(s\) usually reduces the algebra.

Worked examples

Common mistake: losing the minus sign

When \(u=\cos x\), the differential is \(du=-\sin x\,dx\). Dropping this minus sign changes the answer. For example, a mistaken calculation for \(\int\sin^3 x\cos^3 x\,dx\) might be:

This answer has the wrong sign. Differentiating it with \(u=\cos x\) gives \(-\sin^3 x\cos^3 x\), not the original integrand. The corrected substitution includes \(du=-\sin x\,dx\):

Differentiating the corrected expression gives \(\sin x\cos^3 x-\sin x\cos^5 x=\sin x\cos^3 x(1-\cos^2 x)=\sin^3 x\cos^3 x\). A reliable habit is to write the differential substitution explicitly before integrating the polynomial.

Using the method effectively

The essential steps are simple: identify the odd factor to save, convert the remaining even power with the Pythagorean identity, substitute, and integrate the resulting polynomial. The choice of substitution affects the length and appearance of the algebra, but not whether the method works.

The method matters because it turns a trigonometric product into polynomial integration without requiring a special identity beyond \(\sin^2 x+\cos^2 x=1\). It also makes the role of parity explicit: oddness supplies a factor matching the differential, while the remaining even power becomes a polynomial in the substitution variable. For a definite integral on an interval \([a,b]\), the same antiderivative can be evaluated at the endpoints by the Fundamental Theorem of Calculus.