1. Problem: Evaluate the integral $$\int_3^4 \frac{3x}{\sqrt{x-2}} \, dx$$ using the substitution $u = \sqrt{x-2}$. Step 1: Express $x$ in terms of $u$: $$x = u^2 + 2$$. Step 2: Differentiate $x$ to find $dx$: $$dx = 2u \, du$$. Step 3: Change limits: When $x=3$, $u=\sqrt{3-2} = 1$; when $x=4$, $u=\sqrt{4-2} = \sqrt{2}$. Step 4: Substitute into the integral: $$\int_1^{\sqrt{2}} \frac{3(u^2 + 2)}{u} \cdot 2u \, du = \int_1^{\sqrt{2}} 3(u^2 + 2) \cdot 2 \, du = \int_1^{\sqrt{2}} 6(u^2 + 2) \, du$$ Step 5: Simplify and integrate: $$\int_1^{\sqrt{2}} 6u^2 + 12 \, du = 6 \int_1^{\sqrt{2}} u^2 du + 12 \int_1^{\sqrt{2}} du$$ Step 6: Compute integrals: $$6 \left[ \frac{u^3}{3} \right]_1^{\sqrt{2}} + 12 [u]_1^{\sqrt{2}} = 2[u^3]_1^{\sqrt{2}} + 12[u]_1^{\sqrt{2}}$$ Step 7: Evaluate at limits: $$2((\sqrt{2})^3 - 1^3) + 12(\sqrt{2} - 1) = 2(2\sqrt{2} - 1) + 12\sqrt{2} - 12 = 4\sqrt{2} - 2 + 12\sqrt{2} - 12 = 16\sqrt{2} - 14$$ Final answer for 1: $$16\sqrt{2} - 14$$. 2. Problem: Find $$\int \frac{1}{1+e^{-x}} dx$$. Step 1: Recognize that $$\frac{1}{1+e^{-x}} = \frac{e^x}{e^x + 1}$$. Step 2: Use substitution $t = e^x + 1$, then $dt = e^x dx$. Step 3: Rewrite integral: $$\int \frac{e^x}{e^x + 1} dx = \int \frac{1}{t} dt$$ Step 4: Integrate: $$\int \frac{1}{t} dt = \ln|t| + C = \ln(e^x + 1) + C$$ Final answer for 2: $$\ln(e^x + 1) + C$$. 3. Problem: Integrate each with respect to $x$: (a) $$\int \frac{\cos^2 x}{\sin^2 x} dx$$ Step 1: Rewrite as $$\int \cot^2 x \, dx$$. Step 2: Use identity $$\cot^2 x = \csc^2 x - 1$$. Step 3: Integral becomes $$\int (\csc^2 x - 1) dx = \int \csc^2 x dx - \int dx$$ Step 4: Integrate: $$-\cot x - x + C$$ (b) $$\int [3 \sin x + (2 + \sqrt{x})^2] dx$$ Step 1: Expand $(2 + \sqrt{x})^2 = 4 + 4\sqrt{x} + x$ Step 2: Integrate termwise: $$\int 3 \sin x \, dx + \int 4 \, dx + \int 4 x^{1/2} \, dx + \int x \, dx$$ Step 3: Compute each: $$-3 \cos x + 4x + \frac{8}{3} x^{3/2} + \frac{x^2}{2} + C$$ (c) $$\int \frac{4 - 3x \sec^2 x}{x} dx = \int \left( \frac{4}{x} - 3 \sec^2 x \right) dx$$ Step 1: Integrate termwise: $$4 \int \frac{1}{x} dx - 3 \int \sec^2 x dx$$ Step 2: Compute each: $$4 \ln|x| - 3 \tan x + C$$ 4. Problem: Evaluate $$\int_1^{\sqrt{2}} \frac{1}{x^2 \sqrt{4 - x^2}} dx$$ using substitution $x = 2 \cos \theta$. Step 1: Substitute $x = 2 \cos \theta$, then $dx = -2 \sin \theta \, d\theta$. Step 2: Find limits: When $x=1$, $1=2 \cos \theta \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$. When $x=\sqrt{2}$, $\sqrt{2} = 2 \cos \theta \Rightarrow \cos \theta = \frac{\sqrt{2}}{2} \Rightarrow \theta = \frac{\pi}{4}$. Step 3: Substitute and simplify integrand: $$\frac{1}{x^2 \sqrt{4 - x^2}} = \frac{1}{4 \cos^2 \theta \cdot 2 \sin \theta} = \frac{1}{8 \cos^2 \theta \sin \theta}$$ Step 4: Replacing $dx$: $$\int_{\pi/3}^{\pi/4} \frac{1}{8 \cos^2 \theta \sin \theta} \cdot (-2 \sin \theta) d\theta = -\frac{1}{4} \int_{\pi/3}^{\pi/4} \frac{\sin \theta}{\cos^2 \theta \sin \theta} d\theta = -\frac{1}{4} \int_{\pi/3}^{\pi/4} \frac{1}{\cos^2 \theta} d\theta$$ Step 5: Integral of $\sec^2 \theta$ is $\tan \theta$, so: $$-\frac{1}{4} [\tan \theta]_{\pi/3}^{\pi/4} = -\frac{1}{4} (\tan \frac{\pi}{4} - \tan \frac{\pi}{3}) = -\frac{1}{4} (1 - \sqrt{3}) = \frac{\sqrt{3} - 1}{4}$$ Final answer for 4: $$\frac{\sqrt{3} - 1}{4}$$. 5. Problem: (a) Show $$\int_1^3 x^2 \ln x \, dx = 9 \ln 3 - \frac{26}{9}$$. Step 1: Use integration by parts, let: $$u = \ln x \Rightarrow du = \frac{1}{x} dx$$ $$dv = x^2 dx \Rightarrow v = \frac{x^3}{3}$$ Step 2: Apply formula: $$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx$$ Step 3: Compute integral: $$\frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$$ Step 4: Evaluate from 1 to 3: $$\left[ \frac{x^3}{3} \ln x - \frac{x^3}{9} \right]_1^3 = \left( \frac{27}{3} \ln 3 - \frac{27}{9} \right) - \left( \frac{1}{3} \ln 1 - \frac{1}{9} \right) = 9 \ln 3 - 3 - 0 + \frac{1}{9} = 9 \ln 3 - \frac{26}{9}$$ (b) Evaluate $$\int_0^{1/2} \sin^{-1} x \, dx$$ to 2 significant figures. Step 1: Use integration by parts, letting: $$u = \sin^{-1} x, du = \frac{1}{\sqrt{1 - x^2}} dx$$ $$dv = dx, v = x$$ Step 2: Apply formula: $$\int \sin^{-1} x dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$$ Step 3: Use substitution for second integral: Let $w = 1 - x^2$, $dw = -2x dx$, then: $$\int \frac{x}{\sqrt{1-x^2}} dx = -\frac{1}{2} \int w^{-1/2} dw = -\frac{1}{2} \cdot 2 w^{1/2} + C = -\sqrt{1-x^2} + C$$ Step 4: Combine: $$\int \sin^{-1} x dx = x \sin^{-1} x + \sqrt{1-x^2} + C$$ Step 5: Evaluate from 0 to 1/2: $$\left[ x \sin^{-1} x + \sqrt{1-x^2} \right]_0^{1/2} = \frac{1}{2} \sin^{-1} \frac{1}{2} + \sqrt{1-\frac{1}{4}} - (0 + 1)$$ Since $\sin^{-1} \frac{1}{2} = \frac{\pi}{6} \approx 0.5236$: $$= \frac{1}{2} \cdot 0.5236 + \sqrt{\frac{3}{4}} - 1 = 0.2618 + 0.8660 - 1 = 0.1278$$ Rounded to 2 significant figures: $0.13$. (c) Integrate $$\int 3x^2 e^{x/2} dx$$. Step 1: Use substitution $t = \frac{x}{2}$, so $x = 2t$ and $dx = 2 dt$. Step 2: Rewrite integral: $$\int 3(2t)^2 e^t \, 2 dt = \int 3 \cdot 4 t^2 e^t \cdot 2 dt = 24 \int t^2 e^t dt$$ Step 3: Compute $$\int t^2 e^t dt$$ by integration by parts twice: Let $u = t^2$, $dv = e^t dt$ \Rightarrow $du = 2t dt$, $v = e^t$: $$\int t^2 e^t dt = t^2 e^t - \int 2 t e^t dt$$ Step 4: For $$\int 2 t e^t dt$$, use integration by parts again: Let $u = 2 t$, $dv = e^t dt$, $du = 2 dt$, $v = e^t$: $$\int 2 t e^t dt = 2 t e^t - \int 2 e^t dt = 2 t e^t - 2 e^t + C$$ Step 5: Substitute back: $$\int t^2 e^t dt = t^2 e^t - (2 t e^t - 2 e^t) + C = e^t (t^2 - 2 t + 2) + C$$ Step 6: Final answer: $$24 e^t (t^2 - 2 t + 2) + C = 24 e^{x/2} \left( \left( \frac{x}{2} \right)^2 - 2 \cdot \frac{x}{2} + 2 \right) + C = 24 e^{x/2} \left( \frac{x^2}{4} - x + 2 \right) + C$$ Simplify inside: $$24 e^{x/2} \left( \frac{x^2}{4} - x + 2 \right) = 24 e^{x/2} \left( \frac{x^2 - 4x + 8}{4} \right) = 6 e^{x/2} (x^2 - 4x + 8) + C$$ Final answer for 5c: $$6 e^{x/2} (x^2 - 4x + 8) + C$$.