To compute the integral

$\int_{\pi/3}^{\pi/6} \sin^3(3x) \cos^4(3x) \, dx,$

let us proceed step-by-step:

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Step 1: Substitution

We use the substitution: $u = \cos(3x).$

From this substitution, the derivative is: $du = -3\sin(3x) \, dx \quad \text{or} \quad \sin(3x) \, dx = -\frac{1}{3} du.$

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Step 2: Rewrite the limits of integration

When $x = \pi/3$: $u = \cos(3 \cdot \pi/3) = \cos(\pi) = -1.$

When $x = \pi/6$: $u = \cos(3 \cdot \pi/6) = \cos(\pi/2) = 0.$

Thus, the integral limits change from $\pi/3 \to \pi/6$ to $u: -1 \to 0$.

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Step 3: Rewrite the integrand

The integrand $\sin^3(3x)\cos^4(3x)\,dx$ becomes: $\sin^3(3x)\cos^4(3x) \, dx = \sin(3x)(\cos^4(3x))\sin^2(3x) \, dx.$

Using $\sin^2(3x) = 1 - \cos^2(3x) = 1 - u^2$ and substituting $du = -3\sin(3x)\,dx$, the integral becomes: $\int_{\pi/3}^{\pi/6} \sin^3(3x)\cos^4(3x) \, dx = \int_{-1}^{0} u^4(1 - u^2)\left(-\frac{1}{3}\right) du.$

Simplify: $\int_{-1}^{0} \sin^3(3x)\cos^4(3x) \, dx = -\frac{1}{3} \int_{-1}^{0} u^4(1 - u^2) \, du.$

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Step 4: Expand and integrate

Expand $u^4(1 - u^2)$: $u^4(1 - u^2) = u^4 - u^6.$

Thus, the integral becomes: $-\frac{1}{3} \int_{-1}^{0} (u^4 - u^6) \, du = -\frac{1}{3} \left[ \int_{-1}^{0} u^4 \, du - \int_{-1}^{0} u^6 \, du \right].$

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Step 5: Evaluate each term

The formula for $\int u^n \, du$ is: $\int u^n \, du = \frac{u^{n+1}}{n+1}.$

First term:

$\int_{-1}^{0} u^4 \, du = \left[\frac{u^5}{5}\right]_{-1}^{0} = \frac{0^5}{5} - \frac{(-1)^5}{5} = 0 - \left(-\frac{1}{5}\right) = \frac{1}{5}.$

Second term:

$\int_{-1}^{0} u^6 \, du = \left[\frac{u^7}{7}\right]_{-1}^{0} = \frac{0^7}{7} - \frac{(-1)^7}{7} = 0 - \left(-\frac{1}{7}\right) = \frac{1}{7}.$

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Step 6: Combine results

The integral becomes: $-\frac{1}{3} \left( \frac{1}{5} - \frac{1}{7} \right).$

Simplify the terms inside the parentheses: $\frac{1}{5} - \frac{1}{7} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35}.$

Thus, the integral is: $-\frac{1}{3} \cdot \frac{2}{35} = -\frac{2}{105}.$

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Final Answer

$\int_{\pi/3}^{\pi/6} \sin^3(3x) \cos^4(3x) \, dx = -\frac{2}{105}.$

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Would you like a breakdown of any specific step? 😊

Relative Questions

1. What are the rules for changing integration limits during substitution?
2. How is $\sin^2(x) + \cos^2(x) = 1$ used in trigonometric integrals?
3. Why do we include a negative sign in the substitution $du = -3\sin(3x)dx$?
4. What is the formula for definite integrals and its role in this problem?
5. Can the integral be solved without substitution? How?

Tip

Always double-check your limits of integration when making substitutions to avoid sign errors!