Integrating Sine-Even, Cosine-Odd Products

Introduction

When an integrand contains powers of both sine and cosine, the parity of the exponents can point to a convenient substitution. In this tutorial, the sine exponent is even and the cosine exponent is odd. The odd cosine power supplies a factor of \(\cos x\,dx\), which is exactly the differential produced by substituting \(u=\sin x\).

The remaining cosine factors can be rewritten using \(\cos^2 x=1-\sin^2 x\). After the substitution, the integrand is a polynomial in \(u\), so the integration reduces to the power rule.

The Pattern and Its Conditions

Consider integrals of the form

where \(m\) is a nonnegative even integer and \(n\) is a positive odd integer. Write the exponents as \(m=2k\) and \(n=2j+1\), where \(k\) and \(j\) are nonnegative integers. Then separate one cosine factor:

The remaining cosine power is even, so it can be written using the Pythagorean identity:

Now the substitution \(u=\sin x\) has differential \(du=\cos x\,dx\). The factor saved from the odd cosine power supplies that differential. The transformed integrand is \(u^{2k}(1-u^2)^j\), a polynomial in \(u\) because \(j\) is a nonnegative integer.

A General Derivation

The binomial theorem expands \((1-u^2)^j\) as a finite sum. This gives a direct formula for the antiderivative and also explains why the method always produces a polynomial calculation for the stated exponents.

More explicitly, the binomial theorem gives \((1-u^2)^j=\sum_{r=0}^{j}(-1)^r\binom{j}{r}u^{2r}\). Multiplying by \(u^{2k}\) and integrating each term yields the following antiderivative:

Here \(C\) is an arbitrary constant. This formula follows from the substitution above and the power rule; it is not a separate technique. To verify it, differentiate the sum. Differentiating each term contributes a factor of \(\cos x\) from the chain rule:

The third equality uses the binomial theorem in reverse, and the last uses \(1-\sin^2 x=\cos^2 x\). Thus differentiating the proposed antiderivative recovers the original integrand.

Worked Examples

A Common Mistake

A frequent error is to recognize \(u=\sin x\) but forget that the substitution requires the factor \(\cos x\,dx\). For example, the following calculation is incorrect:

The right side has converted all three cosine factors into powers of \(1-u^2\), but it has not retained a cosine factor to supply \(du=\cos x\,dx\). It also uses the wrong power of \(1-u^2\): after saving one cosine, only two cosine factors remain, giving \(\cos^2 x=1-\sin^2 x\). The corrected transformation is

Before substituting, count the cosine factors: save exactly one for the differential, then convert the remaining even power using \(\cos^2 x=1-\sin^2 x\).

When This Method Fits

The method applies directly when the exponents are nonnegative integers, the sine exponent is even, and the cosine exponent is odd. The odd cosine exponent is essential to the simple substitution because it leaves a factor \(\cos x\,dx\) after the other cosine factors are rewritten. If both exponents are even, this particular substitution does not produce a polynomial integral in the same way; that case calls for a different approach.

For the stated parity pattern, the calculation follows a reliable sequence: factor off one cosine, replace each remaining pair of cosine factors by \(1-\sin^2 x\), substitute \(u=\sin x\), expand, and integrate term by term. Differentiation then provides a direct check of the result.