1. **State the problem:** Evaluate the definite integral $$\int_2^3 \cos(x)(2+\sin(x))^5 \, dx$$ where the angles are in radians. 2. **Formula and substitution:** To solve integrals involving compositions like this, use substitution. Let $$u = 2 + \sin(x)$$. 3. **Differentiate substitution:** Then $$\frac{du}{dx} = \cos(x)$$, so $$du = \cos(x) \, dx$$. 4. **Rewrite the integral:** Substitute into the integral: $$\int_2^3 \cos(x)(2+\sin(x))^5 \, dx = \int_{x=2}^{x=3} u^5 \, du$$ Note: When changing limits, calculate new limits for $$u$$: - When $$x=2$$, $$u = 2 + \sin(2)$$ - When $$x=3$$, $$u = 2 + \sin(3)$$ 5. **Evaluate new limits:** $$u_1 = 2 + \sin(2)$$ $$u_2 = 2 + \sin(3)$$ 6. **Integral in terms of $$u$$:** $$\int_{u_1}^{u_2} u^5 \, du$$ 7. **Integrate:** $$\int u^5 \, du = \frac{u^6}{6} + C$$ 8. **Evaluate definite integral:** $$\left. \frac{u^6}{6} \right|_{u_1}^{u_2} = \frac{u_2^6 - u_1^6}{6}$$ 9. **Final answer:** $$\frac{(2 + \sin(3))^6 - (2 + \sin(2))^6}{6}$$ This is the exact value of the integral.