Sine And Cosine Powers By Parity

Why Exponent Parity Guides the Method

An integral involving powers of sine and cosine often becomes simpler as soon as you inspect the exponents. If either exponent is odd, one factor can be saved for a substitution, while the remaining even power can be rewritten using \(\sin^2 x+\cos^2 x=1\). If both exponents are even, there is no single factor that naturally pairs with the derivative of the other function; power-reduction identities are usually the better choice.

This strategy applies to integrals of \(\sin^m x\cos^n x\), where \(m\) and \(n\) are nonnegative integers. The exponents determine the algebraic rewrite; the substitution or identity then makes the integral manageable.

The Parity-Based Decision

For an odd exponent, separate one factor of the function with that exponent. The remaining power is even, so it can be expressed using the square of the other function. The separated factor supplies the differential for a substitution.

For example, if the sine exponent is odd, write \(m=2r+1\), where \(r\) is a nonnegative integer. Then

Now \(u=\cos x\) gives \(du=-\sin x\,dx\), so the integral becomes a polynomial integral in \(u\), possibly multiplied by another integer power of \(u\). If instead the cosine exponent is odd, write \(n=2s+1\) and use \(u=\sin x\).

When both exponents are even, use

These identities follow from \(\cos 2x=\cos^2 x-\sin^2 x\) together with \(\sin^2 x+\cos^2 x=1\). Replacing each squared factor this way reduces the powers and produces sums or products of cosines with multiple angles. Those expressions can be integrated using standard trigonometric identities.

Why the Odd-Power Substitution Works

The substitution is not a guess: it follows from the chain rule in reverse. Suppose the sine exponent is odd, so the integral has the form \(\int \sin^{2r+1}x\cos^n x\,dx\). Separate one \(\sin x\), and rewrite the other sine factors as powers of \(1-\cos^2x\). With \(u=\cos x\) and \(du=-\sin x\,dx\), this gives

The expression on the right is a polynomial in \(u\) when \(r\) and \(n\) are nonnegative integers. After integrating it, substituting \(u=\cos x\) back produces an antiderivative. Differentiating that result applies the chain rule and reverses the substitution, which is why the method works.

When Both Exponents Are Even

If both exponents are even, there is no odd factor to save for a direct substitution of this kind. Instead, reduce the powers using the double-angle identities. For a product, it can also be useful to express the result as a sum of multiple-angle cosines, whose antiderivatives are immediate.

A Common Substitution Mistake

A substitution changes both the variable and the differential. It is not valid to replace \(dx\) by \(du\) without deriving the relationship between them. Consider \(\int\sin^2x\cos^2x\,dx\), and try \(u=\sin x\). The tempting but incorrect calculation is

This maps \(\sin^2x\) to \(u^2\) and \(\cos^2x\) to \(1-u^2\), but it incorrectly treats \(dx\) as though it were \(du\). In fact, \(du=\cos x\,dx\), so the correct transformed differential is \(\cos^2x\,dx=\cos x\,du\). The integrand becomes \(u^2\cos x\,du\), not \(u^2(1-u^2)\,du\). Locally, this can be written as \(\pm u^2\sqrt{1-u^2}\,du\), so the substitution does not directly produce a polynomial in \(u\).

The proposed wrong antiderivative, \(\frac{\sin^3x}{3}-\frac{\sin^5x}{5}\), confirms the error when differentiated:

which is not \(\sin^2x\cos^2x\). For this even-even integral, power reduction provides the appropriate method.

Use the Pattern, Then Verify

Parity is a method-selection tool, not a substitute for checking the algebra. Before integrating, identify which exponents are odd and confirm that the saved factor matches the differential of the proposed substitution. If both exponents are even, reduce powers and keep track of multiple-angle factors when integrating. In every case, differentiating the proposed antiderivative is a direct check that the original integrand has been recovered.