1. **Problem 1a:** Evaluate the integral $$\int \sqrt[3]{1 - x^2} \, dx$$ - This is an integral involving a cube root of a quadratic expression. - We can rewrite the integrand as $$(1 - x^2)^{\frac{1}{3}}$$. - This integral does not have a simple elementary antiderivative, so it typically requires substitution or special functions. 2. **Problem 1b:** Evaluate the integral $$\int 3x^2 (x^3 + 5)^7 \, dx$$ - Use substitution: let $$u = x^3 + 5$$. - Then, $$\frac{du}{dx} = 3x^2$$, so $$du = 3x^2 dx$$. - The integral becomes $$\int u^7 \, du$$. - Integrate: $$\int u^7 \, du = \frac{u^8}{8} + C$$. - Substitute back: $$\frac{(x^3 + 5)^8}{8} + C$$. 3. **Problem 2a:** Evaluate the integral $$\int e^x \sin x \, dx$$ - Use integration by parts or recognize it as a standard integral. - Let $$I = \int e^x \sin x \, dx$$. - Integrate by parts twice or use the formula: $$I = \frac{e^x}{2} (\sin x - \cos x) + C$$. 4. **Problem 2b:** Evaluate the integral $$\int x^2 e^{3x} \, dx$$ - Use integration by parts repeatedly. - Let $$u = x^2$$, $$dv = e^{3x} dx$$. - Then $$du = 2x dx$$, $$v = \frac{e^{3x}}{3}$$. - First integration by parts: $$\int x^2 e^{3x} dx = \frac{x^2 e^{3x}}{3} - \int \frac{2x e^{3x}}{3} dx$$. - Second integration by parts on $$\int x e^{3x} dx$$: Let $$u = x$$, $$dv = e^{3x} dx$$, so $$du = dx$$, $$v = \frac{e^{3x}}{3}$$. - Then: $$\int x e^{3x} dx = \frac{x e^{3x}}{3} - \int \frac{e^{3x}}{3} dx = \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} + C$$. - Substitute back: $$\int x^2 e^{3x} dx = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \left( \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right) + C$$. - Simplify: $$= \frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2 e^{3x}}{27} + C$$. **Final answers:** 1a) $$\int (1 - x^2)^{1/3} dx$$ cannot be expressed in elementary functions simply. 1b) $$\frac{(x^3 + 5)^8}{8} + C$$ 2a) $$\frac{e^x}{2} (\sin x - \cos x) + C$$ 2b) $$\frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2 e^{3x}}{27} + C$$