Integration Evaluation 80E74E

1. **Problem Statement:** Evaluate the integrals: a) $$H = \int_1^e x(\ln x)^2 \, dx$$ b) $$K = \int 3 \sin x \sqrt{1 + \cos x} \, dx$$ 2. **Part (a):** $$H = \int_1^e x(\ln x)^2 \, dx$$ - **Integration technique:** Integration by parts. - **Justification:** The integrand is a product of $x$ and $(\ln x)^2$. Since $\ln x$ is complicated to integrate directly, integration by parts helps reduce the power of $\ln x$. 3. **Step 1:** Let $$u = (\ln x)^2 \quad \Rightarrow \quad du = 2 \ln x \cdot \frac{1}{x} dx = \frac{2 \ln x}{x} dx$$ $$dv = x \, dx \quad \Rightarrow \quad v = \frac{x^2}{2}$$ 4. **Step 2:** Apply integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ So, $$H = \left. \frac{x^2}{2} (\ln x)^2 \right|_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{2 \ln x}{x} dx = \left. \frac{x^2}{2} (\ln x)^2 \right|_1^e - \int_1^e x \ln x \, dx$$ 5. **Step 3:** Evaluate the boundary term: At $x=e$: $$\frac{e^2}{2} (\ln e)^2 = \frac{e^2}{2} \cdot 1^2 = \frac{e^2}{2}$$ At $x=1$: $$\frac{1^2}{2} (\ln 1)^2 = \frac{1}{2} \cdot 0 = 0$$ So, $$H = \frac{e^2}{2} - \int_1^e x \ln x \, dx$$ 6. **Step 4:** Now evaluate $$\int_1^e x \ln x \, dx$$ using integration by parts again. Let $$u = \ln x \quad \Rightarrow \quad du = \frac{1}{x} dx$$ $$dv = x \, dx \quad \Rightarrow \quad v = \frac{x^2}{2}$$ 7. **Step 5:** Apply integration by parts: $$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} dx$$ 8. **Step 6:** Simplify the integral: $$\int \frac{x}{2} dx = \frac{1}{2} \int x dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}$$ So, $$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$$ 9. **Step 7:** Evaluate definite integral from 1 to $e$: At $x=e$: $$\frac{e^2}{2} \cdot 1 - \frac{e^2}{4} = \frac{e^2}{2} - \frac{e^2}{4} = \frac{e^2}{4}$$ At $x=1$: $$\frac{1}{2} \cdot 0 - \frac{1}{4} = -\frac{1}{4}$$ So, $$\int_1^e x \ln x \, dx = \frac{e^2}{4} - \left(-\frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}$$ 10. **Step 8:** Substitute back into $H$: $$H = \frac{e^2}{2} - \frac{e^2 + 1}{4} = \frac{2e^2}{4} - \frac{e^2 + 1}{4} = \frac{2e^2 - e^2 - 1}{4} = \frac{e^2 - 1}{4}$$ --- 11. **Part (b):** $$K = \int 3 \sin x \sqrt{1 + \cos x} \, dx$$ - **Integration technique:** Substitution. - **Justification:** The integrand contains $\sin x$ and $\sqrt{1 + \cos x}$. Using substitution with $u = 1 + \cos x$ simplifies the square root and relates $du$ to $\sin x$. 12. **Step 1:** Let $$u = 1 + \cos x \quad \Rightarrow \quad du = -\sin x \, dx$$ So, $$\sin x \, dx = -du$$ 13. **Step 2:** Rewrite the integral: $$K = \int 3 \sin x \sqrt{1 + \cos x} \, dx = 3 \int \sin x \sqrt{u} \, dx = 3 \int \sqrt{u} (-du) = -3 \int u^{1/2} du$$ 14. **Step 3:** Integrate: $$-3 \int u^{1/2} du = -3 \cdot \frac{u^{3/2}}{\frac{3}{2}} + C = -3 \cdot \frac{2}{3} u^{3/2} + C = -2 u^{3/2} + C$$ 15. **Step 4:** Substitute back $u = 1 + \cos x$: $$K = -2 (1 + \cos x)^{3/2} + C$$ --- **Final answers:** $$H = \frac{e^2 - 1}{4}$$ $$K = -2 (1 + \cos x)^{3/2} + C$$