Evaluate the integral: I = integral sin² 3x sin³ 2x dx
Evaluate the integral: $$ I = \int \sin^2 3x\,\sin^3 2x\,dx. $$
Explanation of Method
Rewriting $\sin^23x$ via the power-reduction identity and $\sin^32x$ via the triple-angle identity $\sin^3\theta=\tfrac14(3\sin\theta-\sin3\theta)$ turns the product into a sum of single sine and cosine terms at various frequencies, which can then be combined pairwise with product-to-sum identities.
- Rewrite $\sin^23x$ and $\sin^32x$ using standard identities.
- Multiply and expand using product-to-sum identities.
- Collect the resulting single-frequency sine terms.
- Integrate term by term.
Step 1 · Rewrite each factor
Using the power-reduction identity $\sin^2\theta=\tfrac{1-\cos2\theta}{2}$ with $\theta=3x$:
$$ \sin^23x = \frac{1-\cos6x}{2}. $$
Using the triple-angle identity $\sin3\theta=3\sin\theta-4\sin^3\theta$, solved for $\sin^3\theta=\tfrac14(3\sin\theta-\sin3\theta)$, with $\theta=2x$:
$$ \sin^32x = \frac14\left(3\sin2x-\sin6x\right). $$
Step 2 · Multiply and expand
Multiplying the two expressions:
$$ \sin^23x\,\sin^32x = \frac{1-\cos6x}{2}\cdot\frac{3\sin2x-\sin6x}{4} = \frac18\left(1-\cos6x\right)\left(3\sin2x-\sin6x\right). $$
Expanding the product:
$$ (1-\cos6x)(3\sin2x-\sin6x) = 3\sin2x - \sin6x - 3\sin2x\cos6x + \sin6x\cos6x. $$
Step 3 · Reduce the remaining products
Using $\sin\alpha\cos\alpha=\tfrac12\sin2\alpha$ with $\alpha=6x$:
$$ \sin6x\cos6x = \frac12\sin12x. $$
Using $\sin P\cos Q=\tfrac12[\sin(P+Q)+\sin(P-Q)]$ with $P=2x$, $Q=6x$:
$$ 3\sin2x\cos6x = \frac32\left[\sin8x+\sin(-4x)\right] = \frac32\left(\sin8x-\sin4x\right). $$
Substituting both back:
$$ (1-\cos6x)(3\sin2x-\sin6x) = 3\sin2x - \sin6x - \frac32\sin8x + \frac32\sin4x + \frac12\sin12x. $$
Step 4 · Collect and integrate
So the integrand becomes:
$$ \sin^23x\,\sin^32x = \frac38\sin2x - \frac18\sin6x + \frac{3}{16}\sin4x - \frac{3}{16}\sin8x + \frac{1}{16}\sin12x. $$
Integrating term by term:
$$ I = -\frac{3}{16}\cos2x + \frac{1}{48}\cos6x - \frac{3}{64}\cos4x + \frac{3}{128}\cos8x - \frac{1}{192}\cos12x + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac38\sin2x$ | $-\tfrac{3}{16}\cos2x$ |
| $-\tfrac18\sin6x$ | $\tfrac{1}{48}\cos6x$ |
| $\tfrac{3}{16}\sin4x$ | $-\tfrac{3}{64}\cos4x$ |
| $-\tfrac{3}{16}\sin8x$ | $\tfrac{3}{128}\cos8x$ |
| $\tfrac{1}{16}\sin12x$ | $-\tfrac{1}{192}\cos12x$ |
Additional Notes
- The integrand is defined for all real $x$.
- Using the triple-angle identity $\sin^3\theta=\tfrac14(3\sin\theta-\sin3\theta)$ instead of the Pythagorean splitting $\sin^3\theta=\sin\theta(1-\cos^2\theta)$ avoids ever having to expand a product like $\cos^22x\cos6x$, which is the step where errors are easy to introduce — every term here comes from a single, clean product-to-sum application.
- The result can be checked by differentiating the final expression: term by term this reconstructs $\tfrac38\sin2x-\tfrac18\sin6x+\tfrac{3}{16}\sin4x-\tfrac{3}{16}\sin8x+\tfrac{1}{16}\sin12x$, which is exactly the expanded integrand from Step 4.
Result
The solution to the integral is: