Evaluate the integrals in Exercises 1–22.
∫₀^(π/6) 3cos⁵(3x) dx

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Recognize that the integral is \( \int_0^{\frac{\pi}{6}} 3 \cos^5(3x) \, dx \). The constant 3 can be factored out: \( 3 \int_0^{\frac{\pi}{6}} \cos^5(3x) \, dx \).

Use the substitution \( u = 3x \), which implies \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). Also, change the limits of integration: when \( x=0 \), \( u=0 \); when \( x=\frac{\pi}{6} \), \( u=\frac{\pi}{2} \).

Rewrite the integral in terms of \( u \): \( 3 \int_0^{\frac{\pi}{6}} \cos^5(3x) \, dx = 3 \int_0^{\frac{\pi}{2}} \cos^5(u) \cdot \frac{du}{3} = \int_0^{\frac{\pi}{2}} \cos^5(u) \, du \).

To integrate \( \cos^5(u) \), express it as \( \cos^4(u) \cos(u) \) and use the identity \( \cos^2(u) = 1 - \sin^2(u) \) to rewrite \( \cos^4(u) = (\cos^2(u))^2 = (1 - \sin^2(u))^2 \).

Use the substitution \( t = \sin(u) \), so \( dt = \cos(u) \, du \). The integral becomes \( \int_0^{\frac{\pi}{2}} \cos^5(u) \, du = \int_0^1 (1 - t^2)^2 \, dt \). Expand the integrand and integrate term-by-term.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration of Trigonometric Functions

This involves finding the integral of functions involving sine, cosine, or other trig functions. Techniques often include using identities to simplify powers or products of trig functions before integrating.

Trigonometric Power-Reduction Formulas

Power-reduction formulas express powers of sine or cosine in terms of functions with lower powers or multiple angles, making integration manageable. For example, cos^5(θ) can be rewritten using these formulas to simplify the integral.

Definite Integration and Limits

Definite integration calculates the exact area under a curve between two bounds. After finding the antiderivative, you evaluate it at the upper and lower limits (here 0 and π/6) and subtract to get the final value.

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