To evaluate the integral

$\int_0^{\pi/12} \sin^3(6x) \, dx,$

we will use trigonometric identities and integration techniques.

Step 1: Simplify $\sin^3(6x)$

Using the identity: $\sin^3(u) = \sin(u)(1 - \cos^2(u)),$ we rewrite: $\sin^3(6x) = \sin(6x) - \sin(6x)\cos^2(6x).$

Step 2: Split the integral

$\int_0^{\pi/12} \sin^3(6x) \, dx = \int_0^{\pi/12} \sin(6x) \, dx - \int_0^{\pi/12} \sin(6x)\cos^2(6x) \, dx.$

Sub-Integral 1: $\int \sin(6x) \, dx$

The integral of $\sin(kx)$ is: $\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx).$ So: $\int_0^{\pi/12} \sin(6x) \, dx = -\frac{1}{6} [\cos(6x)]_0^{\pi/12}.$ Evaluate: $\cos(6 \cdot \pi/12) = \cos(\pi/2) = 0, \quad \cos(6 \cdot 0) = \cos(0) = 1.$ Thus: $\int_0^{\pi/12} \sin(6x) \, dx = -\frac{1}{6} (0 - 1) = \frac{1}{6}.$

Sub-Integral 2: $\int \sin(6x)\cos^2(6x) \, dx$

Use the substitution $u = \cos(6x)$, so: $\frac{du}{dx} = -6 \sin(6x) \quad \text{or} \quad \sin(6x) \, dx = -\frac{1}{6} du.$ The bounds for $u = \cos(6x)$ change as follows:

• When $x = 0$: $u = \cos(6 \cdot 0) = 1$,
• When $x = \pi/12$: $u = \cos(6 \cdot \pi/12) = \cos(\pi/2) = 0$.

The integral becomes: $\int_0^{\pi/12} \sin(6x)\cos^2(6x) \, dx = -\frac{1}{6} \int_1^0 u^2 \, du.$ Simplify: $-\frac{1}{6} \int_1^0 u^2 \, du = \frac{1}{6} \int_0^1 u^2 \, du.$ The integral of $u^2$ is: $\int u^2 \, du = \frac{u^3}{3}.$ Evaluate: $\frac{1}{6} \int_0^1 u^2 \, du = \frac{1}{6} \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{6} \cdot \frac{1}{3} (1^3 - 0^3) = \frac{1}{18}.$

Step 3: Combine results

$\int_0^{\pi/12} \sin^3(6x) \, dx = \frac{1}{6} - \frac{1}{18}.$ Simplify: $\frac{1}{6} = \frac{3}{18}, \quad \frac{3}{18} - \frac{1}{18} = \frac{2}{18} = \frac{1}{9}.$

Final Answer:

$\int_0^{\pi/12} \sin^3(6x) \, dx = \frac{1}{9}.$

Would you like a step-by-step explanation of any part? Here are related questions:

1. What substitution techniques work best for trigonometric integrals?
2. How can trigonometric identities simplify complex powers of sine or cosine?
3. What is the process for changing integration bounds during substitution?
4. How can the integral of $\sin^n(x)$ be generalized?
5. Why does symmetry often simplify definite integrals with trigonometric functions?

Tip: Always verify trigonometric integrals by differentiating the result to ensure