Question

Evaluate the integrals in Exercises 1–22.∫₀^(π/2) sin²(2θ) cos³(2θ) dθ

Accepted Answer

Start by recognizing the integral: $$\$$int_0$$^{\frac{\pi}{2}} \sin^2$(2	heta) \cos^3$(2	heta) \, d	heta. $$Notice that the powers of sine and cosine are both positive integers, which suggests using trigonometric identities or substitution. Use the substitution $$u = 2	heta$$, which implies $$du = 2 \, d	heta or$$ $$d	heta = \frac{du}{2}. $$Also, change the limits of integration accordingly: when $$	heta = 0$$, $$u = 0$$; when $$	heta = \frac{\pi}{2}$$, $$u = \pi. $$Rewrite the integral in terms of $$u$$: $$\$$int_0$$^{\pi} \sin^2$(u) \cos^3$(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \$$int_0$$^{\pi} \sin^2$(u) \cos^3$(u) \, du. $$Express $$\cos^3$(u) as$$ $$\cos(u) \cdot \cos^2$(u)$$ and use the Pythagorean identity $$\cos^2$(u) = 1 - \sin^2$(u) to $$rewrite the integral in terms of $$\sin(u)$$: $$\frac{1}{2} \$$int_0$$^{\pi} \sin^2$(u) \cos(u) (1 - \sin^2$(u)) \, du. $$Use the substitution $$t = \sin(u)$$, so $$dt = \cos(u) \, du. $$The integral becomes $$\frac{1}{2} \int_{t=\sin(0)}^{t=\sin(\pi)} t^2$ (1 - t^2$) \, dt. $$Then, expand the integrand and prepare to integrate term-by-term.

Evaluate the integrals in Exercises 1–22.
∫₀^(π/2) sin²(2θ) cos³(2θ) dθ

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Key Concepts

Trigonometric Integrals

Power-Reducing and Double-Angle Identities

Substitution Method