1. **Problem Statement:** (a) Describe definite and indefinite integration. (b) Prove that $$\int_0^{\frac{\pi}{2}} \frac{1}{1 + 2 \cos x} \, dx = \frac{1}{\sqrt{3}} \ln \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right)$$. (c) Find $$\int x^2 \cos^2 x \, dx$$. 2. **Definitions and Important Rules:** - **Indefinite Integration:** It is the process of finding the antiderivative of a function, represented as $$\int f(x) \, dx = F(x) + C$$, where $$C$$ is the constant of integration. - **Definite Integration:** It calculates the net area under the curve of $$f(x)$$ from $$a$$ to $$b$$, given by $$\int_a^b f(x) \, dx = F(b) - F(a)$$. 3. **Proof of the definite integral:** Given: $$I = \int_0^{\frac{\pi}{2}} \frac{1}{1 + 2 \cos x} \, dx$$. Use the substitution $$t = \tan \frac{x}{2}$$, then: $$\cos x = \frac{1 - t^2}{1 + t^2}$$ and $$dx = \frac{2}{1 + t^2} dt$$. Rewrite the integral: $$I = \int_0^1 \frac{1}{1 + 2 \frac{1 - t^2}{1 + t^2}} \cdot \frac{2}{1 + t^2} dt = \int_0^1 \frac{2}{(1 + t^2) + 2(1 - t^2)} dt$$ Simplify denominator: $$(1 + t^2) + 2(1 - t^2) = 1 + t^2 + 2 - 2 t^2 = 3 - t^2$$ So, $$I = \int_0^1 \frac{2}{3 - t^2} dt = 2 \int_0^1 \frac{1}{3 - t^2} dt$$ Rewrite denominator: $$3 - t^2 = (\sqrt{3})^2 - t^2$$ Use formula: $$\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C$$ Apply it: $$I = 2 \cdot \frac{1}{2 \sqrt{3}} \left[ \ln \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| \right]_0^1 = \frac{1}{\sqrt{3}} \left[ \ln \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) - \ln(1) \right]$$ Since $$\ln(1) = 0$$, $$I = \frac{1}{\sqrt{3}} \ln \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right)$$ This proves the given integral. 4. **Finding $$\int x^2 \cos^2 x \, dx$$:** Use the identity: $$\cos^2 x = \frac{1 + \cos 2x}{2}$$ Rewrite integral: $$\int x^2 \cos^2 x \, dx = \int x^2 \frac{1 + \cos 2x}{2} dx = \frac{1}{2} \int x^2 dx + \frac{1}{2} \int x^2 \cos 2x \, dx$$ Calculate first integral: $$\frac{1}{2} \int x^2 dx = \frac{1}{2} \cdot \frac{x^3}{3} = \frac{x^3}{6}$$ For $$\int x^2 \cos 2x \, dx$$, use integration by parts twice: Let $$u = x^2$$, $$dv = \cos 2x \, dx$$. Then $$du = 2x \, dx$$, $$v = \frac{\sin 2x}{2}$$. So, $$\int x^2 \cos 2x \, dx = \frac{x^2 \sin 2x}{2} - \int 2x \cdot \frac{\sin 2x}{2} dx = \frac{x^2 \sin 2x}{2} - \int x \sin 2x \, dx$$ Now, integrate $$\int x \sin 2x \, dx$$ by parts: Let $$u = x$$, $$dv = \sin 2x \, dx$$. Then $$du = dx$$, $$v = -\frac{\cos 2x}{2}$$. So, $$\int x \sin 2x \, dx = -\frac{x \cos 2x}{2} + \int \frac{\cos 2x}{2} dx = -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C$$ Putting it back: $$\int x^2 \cos 2x \, dx = \frac{x^2 \sin 2x}{2} - \left(-\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} \right) + C = \frac{x^2 \sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C$$ 5. **Final answer:** $$\int x^2 \cos^2 x \, dx = \frac{x^3}{6} + \frac{1}{2} \left( \frac{x^2 \sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} \right) + C = \frac{x^3}{6} + \frac{x^2 \sin 2x}{4} + \frac{x \cos 2x}{4} - \frac{\sin 2x}{8} + C$$