Using Product-to-Sum Formulas
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.2: “Trigonometric Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-2-trigonometric-integrals |
| Textbook used in class | Stewart, Calculus, Section 7.2: “Trigonometric Integrals” |
Opening Scenario
Integrals like $\displaystyle\int \sin(3x)\cos(5x)\,dx$ do not fit the patterns for powers of a single trig function. The integrand is a product of two different trig functions with different arguments. The product-to-sum identities convert such products into sums, and sums of individual trig functions are easy to integrate.
Quick Reference
Product-to-sum formulas:
$$\sin A\cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$$
$$\sin A\sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$$
$$\cos A\cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$$
Procedure: Identify $A$ and $B$, apply the formula, integrate term by term.
Key Concepts
1. Where the Formulas Come From
The product-to-sum identities come directly from the sum-and-difference formulas: $$\sin(A+B) = \sin A\cos B + \cos A\sin B$$ $$\sin(A-B) = \sin A\cos B - \cos A\sin B.$$
Adding these: $\sin(A+B) + \sin(A-B) = 2\sin A\cos B$, so $\sin A\cos B = \dfrac{1}{2}[\sin(A+B) + \sin(A-B)]$.
The other two identities follow from the sum-and-difference formulas for cosine in the same way.
2. After Applying the Formula
Each term in the resulting sum has the form $\sin(cx)$ or $\cos(cx)$ for some constant $c$. These integrate directly: $$\int\sin(cx)\,dx = -\frac{1}{c}\cos(cx) + C, \qquad \int\cos(cx)\,dx = \frac{1}{c}\sin(cx) + C.$$
3. Special Cases
When $A = B$, the formulas simplify:
- $\sin A\cos A = \dfrac{1}{2}\sin(2A)$. (This is just the double-angle formula.)
- $\sin^2 A = \dfrac{1}{2}[1 - \cos(2A)]$.
- $\cos^2 A = \dfrac{1}{2}[1 + \cos(2A)]$.
These are the same power-reducing identities used for even powers of sine and cosine.
“Product-to-sum formulas only apply when both arguments are the same variable without coefficients.” The formulas hold for any angles $A$ and $B$, including expressions like $3x$ and $5x$. The variable inside each sine or cosine can be any expression; $A$ and $B$ label the two full arguments.
Worked Example
Evaluate $\displaystyle\int \sin(3x)\cos(5x)\,dx$.
Apply $\sin A\cos B = \dfrac{1}{2}[\sin(A-B) + \sin(A+B)]$ with $A = 3x$, $B = 5x$:
$\sin(3x)\cos(5x) = \dfrac{1}{2}[\sin(-2x) + \sin(8x)] = \dfrac{1}{2}[-\sin(2x) + \sin(8x)]$.
$$\int\sin(3x)\cos(5x)\,dx = \frac{1}{2}\int[-\sin(2x) + \sin(8x)]\,dx$$ $$= \frac{1}{2}\left[\frac{\cos(2x)}{2} - \frac{\cos(8x)}{8}\right] + C = \frac{\cos(2x)}{4} - \frac{\cos(8x)}{16} + C.$$
Boxed answer: $\dfrac{\cos(2x)}{4} - \dfrac{\cos(8x)}{16} + C$.
Evaluate $\displaystyle\int \cos(4x)\cos(6x)\,dx$.
$\cos A\cos B = \dfrac{1}{2}[\cos(A-B) + \cos(A+B)]$ with $A = 4x$, $B = 6x$:
$\cos(4x)\cos(6x) = \dfrac{1}{2}[\cos(-2x) + \cos(10x)] = \dfrac{1}{2}[\cos(2x) + \cos(10x)]$.
$$\int\cos(4x)\cos(6x)\,dx = \frac{1}{2}\left[\frac{\sin(2x)}{2} + \frac{\sin(10x)}{10}\right] + C = \frac{\sin(2x)}{4} + \frac{\sin(10x)}{20} + C.$$
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the wrong formula | Applying the $\sin A\sin B$ formula to $\sin A\cos B$ | Check which combination is present: $\sin\cdot\cos$, $\sin\cdot\sin$, or $\cos\cdot\cos$ |
| Forgetting the factor of $\frac{1}{2}$ | Writing $\sin A\cos B = \sin(A-B) + \sin(A+B)$ | The formula has a factor $\frac{1}{2}$: $\sin A\cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$ |
| Sign error from $\sin(-\theta) = -\sin\theta$ | Leaving $\sin(-2x)$ in the integral without converting | $\sin(-2x) = -\sin(2x)$; the negative sign must be applied before integrating |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Evaluate $\displaystyle\int \sin(x)\sin(3x)\,dx$.
Show answer
$\sin A\sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$, $A = x$, $B = 3x$:
$\sin x\sin(3x) = \frac{1}{2}[\cos(-2x) - \cos(4x)] = \frac{1}{2}[\cos(2x) - \cos(4x)]$.
$\displaystyle\int = \frac{1}{2}\left[\frac{\sin(2x)}{2} - \frac{\sin(4x)}{4}\right] + C = \frac{\sin(2x)}{4} - \frac{\sin(4x)}{8} + C$.
Level 2 -- Checking the Special Case
Problem 2. Evaluate $\displaystyle\int \sin(5x)\cos(5x)\,dx$ using a product-to-sum formula, then verify using the substitution $u = \sin(5x)$.
Show answer
Product-to-sum: $A = B = 5x$, $\sin A\cos B = \frac{1}{2}\sin(2A)$, so $\sin(5x)\cos(5x) = \frac{1}{2}\sin(10x)$.
$\displaystyle\int\frac{1}{2}\sin(10x)\,dx = -\frac{\cos(10x)}{20} + C$.
Substitution check: $u = \sin(5x)$, $du = 5\cos(5x)\,dx$, so $\sin(5x)\cos(5x)\,dx = \frac{u\,du}{5}$.
$\displaystyle\int\frac{u}{5}\,du = \frac{u^2}{10} + C = \frac{\sin^2(5x)}{10} + C$.
Are the answers the same? Yes, since $\cos(10x) = 1 - 2\sin^2(5x)$, so $-\frac{\cos(10x)}{20} = -\frac{1-2\sin^2(5x)}{20} = \frac{\sin^2(5x)}{10} - \frac{1}{20}$. The $-\frac{1}{20}$ is absorbed into $C$.
Mastery Checklist
Mental Model
A product of two trig functions with different arguments is hard to integrate directly because there is no substitution that simplifies both factors simultaneously. The product-to-sum formulas trade the product for a sum: instead of integrating one complicated thing, you integrate two simple things. This is a common theme in integration -- when an integrand is a product, try to convert it to a sum.
Connections
Looking back
- Sum-and-difference formulas: The product-to-sum identities are derived from these.
- Power-reducing identities (Section 7.2): The special case $A = B$ gives the power-reducing formulas used for even powers of sine and cosine.
Looking ahead
- Fourier analysis: Products of sines and cosines with integer multiples of a base frequency are the building blocks of Fourier series, where these integrals appear as the key orthogonality integrals.
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