1. **Problem:** Evaluate the integral $$\int (1 - 2x)^3 \, dx$$ Step 1: Use substitution. Let $$u = 1 - 2x$$ then $$du = -2 \, dx$$ or $$dx = -\frac{du}{2}$$. Step 2: Rewrite the integral as $$\int u^3 \left(-\frac{1}{2}\right) du = -\frac{1}{2} \int u^3 du$$. Step 3: Integrate $$u^3$$: $$\int u^3 du = \frac{u^4}{4} + C$$. Step 4: Substitute back $$u = 1 - 2x$$: $$ -\frac{1}{2} \cdot \frac{(1 - 2x)^4}{4} + C = -\frac{(1 - 2x)^4}{8} + C $$. 2. **Problem:** Evaluate $$\int \sin(2\pi x + 7) \, dx$$ Step 1: Substitute $$u = 2\pi x + 7$$, then $$du = 2\pi dx$$, so $$dx = \frac{du}{2\pi}$$. Step 2: Rewrite integral as $$\int \sin u \cdot \frac{1}{2\pi} du = \frac{1}{2\pi} \int \sin u \, du$$. Step 3: Integrate $$\sin u$$: $$\int \sin u \, du = -\cos u + C$$. Step 4: Substitute back $$\frac{-\cos(2\pi x + 7)}{2\pi} + C$$. 3. **Problem:** Evaluate $$\int \cos(3x - 7) \, dx$$ Step 1: Substitute $$u = 3x - 7$$, then $$du = 3 \, dx$$, so $$dx = \frac{du}{3}$$. Step 2: Rewrite integral as $$\int \cos u \cdot \frac{1}{3} du = \frac{1}{3} \int \cos u \, du$$. Step 3: Integrate $$\cos u$$: $$\int \cos u \, du = \sin u + C$$. Step 4: Substitute back $$\frac{\sin(3x - 7)}{3} + C$$. 4. **Problem:** Evaluate $$\int 3^{5x-2} \, dx$$ Step 1: Substitute $$u = 5x - 2$$, then $$du = 5 dx$$, so $$dx = \frac{du}{5}$$. Step 2: Rewrite integral as $$\int 3^u \cdot \frac{1}{5} du = \frac{1}{5} \int 3^u du$$. Step 3: Recall $$\int a^u du = \frac{a^u}{\ln a} + C$$ for $$a > 0$$. Step 4: Apply formula: $$\frac{1}{5} \cdot \frac{3^u}{\ln 3} + C = \frac{3^{5x - 2}}{5 \ln 3} + C$$. 5. **Problem:** Evaluate $$\int \frac{1}{7x - 6} \, dx$$ Step 1: Substitute $$u = 7x - 6$$, then $$du = 7 dx$$, so $$dx = \frac{du}{7}$$. Step 2: Rewrite integral: $$\int \frac{1}{u} \cdot \frac{1}{7} du = \frac{1}{7} \int \frac{1}{u} du$$. Step 3: Integrate: $$\int \frac{1}{u} du = \ln |u| + C$$. Step 4: Substitute back: $$\frac{1}{7} \ln |7x - 6| + C$$. 6. **Problem:** Evaluate $$\int \frac{\sin 2x}{\sin x} \, dx$$ Step 1: Use the double angle formula for sine: $$\sin 2x = 2 \sin x \cos x$$. Step 2: Substitute: $$\int \frac{2 \sin x \cos x}{\sin x} dx = \int 2 \cos x \, dx$$. Step 3: Simplify integral: $$2 \int \cos x \, dx = 2 \sin x + C$$. 7. **Problem:** Evaluate $$\int \sec^2(2x + 3) \, dx$$ Step 1: Substitute $$u = 2x + 3$$, then $$du = 2 dx$$, so $$dx = \frac{du}{2}$$. Step 2: Rewrite integral: $$\int \sec^2 u \cdot \frac{1}{2} du = \frac{1}{2} \int \sec^2 u \, du$$. Step 3: Integrate: $$\int \sec^2 u \, du = \tan u + C$$. Step 4: Substitute back: $$\frac{1}{2} \tan(2x + 3) + C$$. 8. **Problem:** Evaluate $$\int e^{7x - 3} \, dx$$ Step 1: Substitute $$u = 7x - 3$$, then $$du = 7 dx$$, so $$dx = \frac{du}{7}$$. Step 2: Rewrite integral: $$\int e^u \cdot \frac{1}{7} du = \frac{1}{7} \int e^u du$$. Step 3: Integrate: $$\int e^u du = e^u + C$$. Step 4: Substitute back: $$\frac{e^{7x - 3}}{7} + C$$. 9. **Problem:** Evaluate $$\int (1 + \tan^2 2\pi x) \, dx$$ Step 1: Recall the identity: $$1 + \tan^2 \theta = \sec^2 \theta$$. Step 2: So the integral is: $$\int \sec^2 2\pi x \, dx$$. Step 3: Substitute $$u = 2\pi x$$, then $$du = 2\pi dx$$, so $$dx = \frac{du}{2\pi}$$. Step 4: Rewrite integral: $$\int \sec^2 u \cdot \frac{1}{2\pi} du = \frac{1}{2\pi} \int \sec^2 u \, du$$. Step 5: Integrate: $$\int \sec^2 u \, du = \tan u + C$$. Step 6: Substitute back: $$\frac{\tan(2\pi x)}{2\pi} + C$$. **Final answers:** (a) $$-\frac{(1 - 2x)^4}{8} + C$$ (b) $$-\frac{\cos(2\pi x + 7)}{2\pi} + C$$ (c) $$\frac{\sin(3x - 7)}{3} + C$$ (d) $$\frac{3^{5x - 2}}{5 \ln 3} + C$$ (e) $$\frac{1}{7} \ln |7x - 6| + C$$ (f) $$2 \sin x + C$$ (g) $$\frac{1}{2} \tan(2x + 3) + C$$ (h) $$\frac{e^{7x - 3}}{7} + C$$ (i) $$\frac{\tan(2\pi x)}{2\pi} + C$$