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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