1. Problem: Evaluate $$\int_0^4 x^3 \sqrt{4x - x^2} \, dx$$ Step 1: Rewrite the integrand: $$\sqrt{4x - x^2} = \sqrt{-(x^2 -4x)} = \sqrt{4^2 - (x-2)^2}$$ but it's simpler to use substitution. Step 2: Let $$t = 4x - x^2 = x(4-x)$$, then $$dt = (4 - 2x) dx$$. Alternatively, use substitution $$x = 4\sin^2 u$$ to simplify the square root. Step 3: With the substitution $$x = 4\sin^2 u$$, $$dx = 8 \sin u \cos u du$$. Step 4: Then $$4x - x^2 = 16 \sin^2 u - 16 \sin^4 u = 16 \sin^2 u (1- \sin^2 u) = 16 \sin^2 u \cos^2 u$$. So $$\sqrt{4x - x^2} = 4 \sin u \cos u$$. Step 5: Substitute all into the integral: $$\int_0^{4} x^3 \sqrt{4x - x^2} dx = \int_0^{\pi/2} (4 \sin^2 u)^3 (4 \sin u \cos u) (8 \sin u \cos u) du$$ Step 6: Simplify the powers and constants: $$= \int_0^{\pi/2} 64 \sin^6 u \cdot 4 \sin u \cos u \cdot 8 \sin u \cos u du = \int_0^{\pi/2} 64 \times 4 \times 8 \sin^{6+1+1} u \cos^{1+1} u du$$ $$= 2048 \int_0^{\pi/2} \sin^{8} u \cos^{2} u du$$ Step 7: Use Beta function or reduction formulas: $$I = \int_0^{\pi/2} \sin^{m} u \cos^{n} u du = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$$ where $$B$$ is Beta function. Step 8: Here $$m=8$$, $$n=2$$: $$I = \frac{1}{2} B\left(\frac{9}{2}, \frac{3}{2}\right) = \frac{1}{2} \frac{\Gamma(9/2) \Gamma(3/2)}{\Gamma(6)}$$ Step 9: Use Gamma function values: $$\Gamma\left(\frac{9}{2}\right) = \frac{105 \sqrt{\pi}}{16}, \quad \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}, \quad \Gamma(6) = 120$$ Step 10: Compute: $$I = \frac{1}{2} \times \frac{(105 \sqrt{\pi}/16)(\sqrt{\pi}/2)}{120} = \frac{1}{2} \times \frac{105 \pi}{32 \times 120} = \frac{105 \pi}{7680}$$ Step 11: Multiply by constant 2048: $$\int_0^4 x^3 \sqrt{4x - x^2} dx = 2048 \times \frac{105 \pi}{7680} = \frac{2048 \times 105 \pi}{7680} = 28 \pi$$ --- 2. Problem: Evaluate $$\int_0^{\pi/6} \cos^4(3\phi) \sin^3(6\phi) d\phi$$ Step 1: Use substitution: Let $$t=6\phi$$, then $$d\phi = \frac{dt}{6}$$ and when $$\phi=0, t=0$$; when $$\phi=\frac{\pi}{6}, t=\pi$$ Step 2: Rewrite integral: $$\int_0^{\pi/6} \cos^4(3\phi) \sin^3(6\phi) d\phi = \int_0^{\pi} \cos^4\left(\frac{t}{2}\right) \sin^3 t \frac{dt}{6} = \frac{1}{6} \int_0^{\pi} \cos^4\left(\frac{t}{2}\right) \sin^3 t dt$$ Step 3: Use $$\sin^3 t = \sin t \cdot \sin^2 t = \sin t (1 - \cos^2 t)$$ Step 4: Substitute: Let $$u = \cos t$$, then $$du = -\sin t dt$$, so $$\sin t dt = -du$$ Bounds: $$t=0 \to u=1$$, $$t=\pi \to u = -1$$ Step 5: Integral becomes: $$\frac{1}{6} \int_{t=0}^{\pi} \cos^4\left(\frac{t}{2}\right) \sin t (1 - \cos^2 t) dt = \frac{1}{6} \int_{u=1}^{-1} \cos^4\left(\frac{t}{2}\right) (1-u^2)(-du)$$ Step 6: Express $$\cos^4\left(\frac{t}{2}\right)$$ in terms of $$u=\cos t$$. Recall $$\cos t = 2 \cos^2 \frac{t}{2} - 1$$ so $$\cos^2 \frac{t}{2} = \frac{1+u}{2}$$ Thus $$\cos^4 \frac{t}{2} = \left(\frac{1+u}{2}\right)^2 = \frac{(1+u)^2}{4}$$ Step 7: Substituting back: $$\frac{1}{6} \int_1^{-1} \frac{(1+u)^2}{4} (1 - u^2)(-du) = \frac{1}{24} \int_{-1}^{1} (1+u)^2 (1-u^2) du$$ Step 8: Expand integrand: $$(1+u)^2 = 1 + 2u + u^2$$ $$(1 - u^2) = 1 - u^2$$ Multiply: $$(1 + 2u + u^2)(1 - u^2) = (1)(1 - u^2) + 2u(1 - u^2) + u^2(1 - u^2) = 1 - u^2 + 2u - 2u^3 + u^2 - u^4 = 1 + 2u - 2u^3 - u^4$$ Step 9: Integral becomes: $$\frac{1}{24} \int_{-1}^1 (1 + 2u - 2u^3 - u^4) du = \frac{1}{24} \left[ \int_{-1}^1 1 du + 2 \int_{-1}^1 u du - 2 \int_{-1}^1 u^3 du - \int_{-1}^1 u^4 du \right]$$ Step 10: Calculate each integral: $$\int_{-1}^1 1 du = 2$$ $$\int_{-1}^1 u du = 0$$ (odd function) $$\int_{-1}^1 u^3 du = 0$$ (odd function) $$\int_{-1}^1 u^4 du = \frac{2}{5}$$ Step 11: Substitute values: $$\frac{1}{24} (2 + 0 - 0 - \frac{2}{5}) = \frac{1}{24} \left(2 - \frac{2}{5}\right) = \frac{1}{24} \times \frac{10 - 2}{5} = \frac{8}{120} = \frac{1}{15}$$ --- 3. Problem: Evaluate $$\int_0^{3\pi/2} \cos^5 \left(\frac{\theta}{3}\right) d\theta$$ Step 1: Let $$t = \frac{\theta}{3}$$, then $$d\theta = 3 dt$$ Change limits: when $$\theta = 0, t=0$$; when $$\theta = \frac{3\pi}{2}, t= \frac{\pi}{2}$$ Step 2: Integral becomes: $$\int_0^{3\pi/2} \cos^5 \left(\frac{\theta}{3} \right) d\theta = 3 \int_0^{\pi/2} \cos^5 t dt$$ Step 3: Use power-reduction formula or standard integral: $$\int \cos^n x dx$$ for odd $$n$$: Separate one cosine and write: $$\cos^5 t = \cos^4 t \cos t = (\cos^2 t)^2 \cos t = (1 - \sin^2 t)^2 \cos t$$ Step 4: Let $$u = \sin t$$, $$du = \cos t dt$$ Step 5: Then integral becomes: $$3 \int_0^{\pi/2} (1 - u^2)^2 du$$ with $$u$$ from $$0$$ to $$1$$ Step 6: Expand: $$(1 - u^2)^2 = 1 - 2 u^2 + u^4$$ Step 7: Integral evaluates to: $$3 \int_0^{1} (1 - 2 u^2 + u^4) du = 3 \left[ u - \frac{2 u^3}{3} + \frac{u^5}{5} \right]_0^{1} = 3 \left(1 - \frac{2}{3} + \frac{1}{5}\right)$$ Step 8: Compute inside: $$1 - \frac{2}{3} + \frac{1}{5} = \frac{15}{15} - \frac{10}{15} + \frac{3}{15} = \frac{8}{15}$$ Step 9: Multiply by 3: $$3 \times \frac{8}{15} = \frac{24}{15} = \frac{8}{5}$$ --- 4. Problem: Evaluate $$\int_0^{\pi/4} (\cos 2\theta)^{3/2} \cos \theta d\theta$$ Step 1: Use substitution: Let $$t = 2\theta$$, then $$d\theta = \frac{dt}{2}$$ Bounds: when $$\theta=0, t=0$$; when $$\theta=\frac{\pi}{4}, t=\frac{\pi}{2}$$ Step 2: Integral becomes: $$\int_0^{\pi/4} (\cos 2\theta)^{3/2} \cos \theta d\theta = \int_0^{\pi/2} (\cos t)^{3/2} \cos \left(\frac{t}{2}\right) \frac{dt}{2} = \frac{1}{2} \int_0^{\pi/2} (\cos t)^{3/2} \cos \frac{t}{2} dt$$ Step 3: Express in simpler form or use substitution for $$\cos \frac{t}{2}$$ but likely requires advanced techniques or numerical methods. No simple closed form; the integral remains as is. --- 5. Problem: Evaluate $$\int_0^{\pi} \sin^4 \left(\frac{\theta}{2}\right) \cos^3 \left(\frac{\theta}{2}\right) d\theta$$ Step 1: Substitute $$u = \frac{\theta}{2}$$ so $$d\theta = 2 du$$ Bounds: $$\theta=0 \to u=0$$, $$\theta=\pi \to u = \frac{\pi}{2}$$ Step 2: Integral becomes: $$2 \int_0^{\pi/2} \sin^4 u \cos^3 u du$$ Step 3: Let $$I = \int_0^{\pi/2} \sin^4 u \cos^3 u du$$ Step 4: Use substitution $$t = \sin u$$, $$dt = \cos u du$$ Rewrite: $$I = \int_0^1 t^4 (1 - t^2)^{1} (1-t^2)^{1/2} dt$$ but better to write: $$I = \int_0^{\pi/2} \sin^4 u (\cos^2 u) \cos u du = \int_0^{\pi/2} \sin^4 u (1 - \sin^2 u) \cos u du$$ since $$\cos^3 u = \cos u \times \cos^2 u$$ Step 5: With $$t = \sin u$$, $$dt = \cos u du$$, $$I = \int_0^1 t^4 (1 - t^2) dt = \int_0^1 (t^4 - t^6) dt$$ Step 6: Integrate: $$\left[ \frac{t^5}{5} - \frac{t^7}{7} \right]_0^1 = \frac{1}{5} - \frac{1}{7} = \frac{7 - 5}{35} = \frac{2}{35}$$ Step 7: Multiply by 2: $$2 \times \frac{2}{35} = \frac{4}{35}$$ --- 6. Problem: Evaluate $$\int_0^1 x^2 (1-x^2)^{3/2} dx$$ Step 1: Use substitution $$t = x^2$$, so $$dt = 2x dx$$, $$x dx = \frac{dt}{2}$$ Rewrite integral: $$\int_0^1 x^2 (1 - x^2)^{3/2} dx = \int_0^1 x^2 (1 - t)^{3/2} dx$$ Since $$t = x^2$$, $$x = \sqrt{t}$$, so: $$dx = \frac{dt}{2 \sqrt{t}}$$ Step 2: Substitute for $$x^2 dx$$: $$x^2 dx = t dx = t \cdot \frac{dt}{2 \sqrt{t}} = \frac{t}{2 \sqrt{t}} dt = \frac{\sqrt{t}}{2} dt$$ Step 3: Integral becomes: $$\int_0^1 (1 - t)^{3/2} \frac{\sqrt{t}}{2} dt = \frac{1}{2} \int_0^1 t^{1/2} (1 - t)^{3/2} dt$$ Step 4: Recognize this as Beta function integral: $$B(p,q) = \int_0^1 t^{p-1} (1 - t)^{q-1} dt$$ Here, $$p = \frac{3}{2}$$, $$q=\frac{5}{2}$$ Step 5: Use Beta function relation: $$\int_0^1 t^{p-1} (1 - t)^{q-1} dt = B(p,q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$ Step 6: Compute Beta function: $$I = \frac{1}{2} B\left(\frac{3}{2}, \frac{5}{2}\right) = \frac{1}{2} \cdot \frac{\Gamma(\frac{3}{2}) \Gamma(\frac{5}{2})}{\Gamma(4)}$$ Step 7: Use Gamma values: $$\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$$ $$\Gamma\left(\frac{5}{2}\right) = \frac{3 \sqrt{\pi}}{4}$$ $$\Gamma(4) = 3! = 6$$ Step 8: Substitute: $$I = \frac{1}{2} \times \frac{(\sqrt{\pi}/2)(3 \sqrt{\pi}/4)}{6} = \frac{1}{2} \times \frac{3 \pi}{8 \times 6} = \frac{3 \pi}{96} = \frac{\pi}{32}$$ --- Final answers: 1) $$28 \pi$$ 2) $$\frac{1}{15}$$ 3) $$\frac{8}{5}$$ 4) Integral equals $$\frac{1}{2} \int_0^{\pi/2} (\cos t)^{3/2} \cos \frac{t}{2} dt$$ (complex, closed form not trivial) 5) $$\frac{4}{35}$$ 6) $$\frac{\pi}{32}$$