Integrating the Product of Sine and Cosine: ∫ sin x cos 2x dx
The integral of a product of trigonometric functions often appears in calculus courses, physics problems, and engineering applications. One particularly common example is
[ \int \sin x , \cos 2x , dx . ]
Although the integrand seems simple, directly integrating it is not straightforward because the integrand is a product of two different trigonometric functions. This article walks through multiple strategies—product-to-sum identities, substitution, and integration by parts—to evaluate the integral, explains why each method works, and discusses the broader context of integrating products of sines and cosines.
Introduction
When faced with an integral like (\int \sin x \cos 2x , dx), the first instinct is to look for a direct antiderivative. Still, the product of two trigonometric functions of different arguments resists simple antiderivatives in the usual elementary function form. Instead, we can transform the product into a sum (or difference) of simpler trigonometric functions using product‑to‑sum identities. Once transformed, the integral becomes a linear combination of standard integrals that are easy to evaluate.
Step 1: Apply a Product-to-Sum Identity
The product-to-sum formulas for sine and cosine are:
- (\sin A \cos B = \frac{1}{2}\bigl[\sin(A+B) + \sin(A-B)\bigr]).
Setting (A = x) and (B = 2x) gives
[ \sin x \cos 2x = \frac{1}{2}\bigl[\sin(3x) + \sin(-x)\bigr]. ]
Since (\sin(-x) = -\sin x), this simplifies to
[ \sin x \cos 2x = \frac{1}{2}\bigl[\sin 3x - \sin x\bigr]. ]
Thus, the integral becomes
[ \int \sin x \cos 2x , dx = \frac{1}{2}\int \sin 3x , dx - \frac{1}{2}\int \sin x , dx. ]
Step 2: Integrate Each Term Separately
Both integrals are standard:
- (\int \sin kx , dx = -\frac{1}{k}\cos kx + C).
Applying this rule:
- (\displaystyle \int \sin 3x , dx = -\frac{1}{3}\cos 3x + C_1).
- (\displaystyle \int \sin x , dx = -\cos x + C_2).
Substituting back:
[ \frac{1}{2}\left(-\frac{1}{3}\cos 3x\right) - \frac{1}{2}\left(-\cos x\right) + C = -\frac{1}{6}\cos 3x + \frac{1}{2}\cos x + C. ]
So the antiderivative is
[ \boxed{\displaystyle \int \sin x \cos 2x , dx = \frac{1}{2}\cos x - \frac{1}{6}\cos 3x + C.} ]
Alternative Approach 1: Substitution
Sometimes a substitution can simplify the integrand directly. Observe that
[ \frac{d}{dx}\bigl(\cos 2x\bigr) = -2\sin 2x, ] but our integrand involves (\sin x), not (\sin 2x). That said, we can use the identity (\cos 2x = 1 - 2\sin^2 x) or (\cos 2x = 2\cos^2 x - 1). Let’s try the substitution (u = \cos x). Then (du = -\sin x , dx), so (-du = \sin x , dx).
[ \int \sin x \cos 2x , dx = -\int \cos 2x , du. ]
Now express (\cos 2x) in terms of (u):
[ \cos 2x = 2\cos^2 x - 1 = 2u^2 - 1. ]
Thus,
[ -\int (2u^2 - 1) , du = -\left(\frac{2}{3}u^3 - u\right) + C = -\frac{2}{3}u^3 + u + C. ]
Re-substitute (u = \cos x):
[ -\frac{2}{3}\cos^3 x + \cos x + C. ]
Differentiating this expression confirms it equals (\sin x \cos 2x). Notice that this result is algebraically equivalent to the product‑to‑sum result:
[ -\frac{2}{3}\cos^3 x + \cos x = \frac{1}{2}\cos x - \frac{1}{6}\cos 3x, ]
using the triple‑angle identity (\cos 3x = 4\cos^3 x - 3\cos x).
Alternative Approach 2: Integration by Parts
Integration by parts is another systematic technique:
[ \int u,dv = uv - \int v,du. ]
Choose (u = \cos 2x) and (dv = \sin x , dx). Then
- (du = -2\sin 2x , dx),
- (v = -\cos x).
Apply the formula:
[ \int \sin x \cos 2x , dx = (-\cos x)(\cos 2x) - \int (-\cos x)(-2\sin 2x), dx = -\cos x \cos 2x - 2\int \cos x \sin 2x , dx. ]
The remaining integral (\int \cos x \sin 2x , dx) can be simplified using the product‑to‑sum identity again:
[ \cos x \sin 2x = \frac{1}{2}\bigl[\sin(3x) - \sin(-x)\bigr] = \frac{1}{2}\bigl[\sin 3x + \sin x\bigr]. ]
Thus,
[
- 2\int \cos x \sin 2x , dx = -\int \bigl[\sin 3x + \sin x\bigr] , dx = \frac{1}{3}\cos 3x + \cos x + C. ]
Combine with the first term:
[ -\cos x \cos 2x + \frac{1}{3}\cos 3x + \cos x + C. ]
Using the identity (\cos x \cos 2x = \frac{1}{2}\bigl[\cos(3x) + \cos(-x)\bigr] = \frac{1}{2}\bigl[\cos 3x + \cos x\bigr]), we simplify to
[ -\frac{1}{2}\cos 3x - \frac{1}{2}\cos x + \frac{1}{3}\cos 3x + \cos x = \frac{1}{2}\cos x - \frac{1}{6}\cos 3x + C, ]
the same antiderivative obtained earlier.
Scientific Explanation: Why the Product-to-Sum Works
The product-to-sum identities arise from the addition formulas for sine and cosine:
[ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B, ] [ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B. ]
By adding or subtracting these equations appropriately, we can solve for (\sin A \cos B) or (\cos A \sin B). This transformation turns a product of two trigonometric functions into a sum of single trigonometric functions with arguments that are linear combinations of the original angles. Since the antiderivatives of (\sin(kx)) and (\cos(kx)) are elementary, the integral becomes trivial.
Generalization: Integrals of the Form ∫ sin mx cos nx dx
The same strategy applies to any integral of the form
[ \int \sin(mx)\cos(nx),dx, ]
where (m) and (n) are constants. Using the product-to-sum identity:
[ \sin(mx)\cos(nx) = \frac{1}{2}\bigl[\sin((m+n)x) + \sin((m-n)x)\bigr]. ]
Then
[ \int \sin(mx)\cos(nx),dx = \frac{1}{2}\left(-\frac{1}{m+n}\cos((m+n)x) - \frac{1}{m-n}\cos((m-n)x)\right) + C, ]
provided (m \neq n). If (m = n), the identity reduces to (\sin(mx)\cos(mx) = \frac{1}{2}\sin(2mx)), and the integral follows accordingly Worth keeping that in mind..
FAQ
| Question | Answer |
|---|---|
| *Can I integrate sin x cos 2x by using a simple substitution u = sin x?Thus, this substitution alone is insufficient. * | Substituting (u = \sin x) gives (du = \cos x,dx), but the integrand contains (\cos 2x), which is not expressed in terms of (u). Still, * |
| *What if I want to integrate sin x cos x?Here's the thing — * | Use the identity (\sin x \cos x = \frac{1}{2}\sin 2x) and integrate (\frac{1}{2}\int \sin 2x,dx = -\frac{1}{4}\cos 2x + C). |
| *Is there a shortcut for integrals involving sin x cos nx where n is an integer?Now, | |
| *How do I check if my antiderivative is correct? * | Differentiate your result; if you recover the original integrand, the antiderivative is correct. |
Conclusion
Evaluating (\displaystyle \int \sin x \cos 2x , dx) illustrates a powerful technique in integral calculus: converting products of trigonometric functions into sums via product-to-sum identities. This transformation reduces the problem to integrating elementary sine and cosine functions, yielding the antiderivative
Counterintuitive, but true.
[ \boxed{\frac{1}{2}\cos x - \frac{1}{6}\cos 3x + C}. ]
Alternative methods—substitution or integration by parts—lead to the same result, offering flexibility depending on the integrand’s structure. Mastering these techniques not only solves this particular integral but also equips you to tackle a wide array of trigonometric integrals encountered in mathematics, physics, and engineering.
