1. Problem (b): Evaluate \(\int \frac{3}{\sin \theta - 3 \cos \theta - 1} \, d\theta\) using the substitution \(t = \tan(\frac{\theta}{2})\), and then find \(\int \frac{7 \sin \theta - \cos \theta + 2}{\sin \theta - 3 \cos \theta - 1} \, d\theta\). 2. Recall the Weierstrass substitution: $$\sin \theta = \frac{2t}{1+t^2}, \quad \cos \theta = \frac{1 - t^2}{1+t^2}, \quad d\theta = \frac{2}{1+t^2} dt.$$ 3. Substitute into denominator of the first integral: $$\sin \theta - 3 \cos \theta - 1 = \frac{2t}{1+t^2} - 3 \cdot \frac{1 - t^2}{1+t^2} - 1 = \frac{2t - 3(1 - t^2) - (1+t^2)}{1+t^2} = \frac{2t - 3 + 3t^2 - 1 - t^2}{1+t^2} = \frac{2t + 2t^2 - 4}{1+t^2}.$$ 4. So the first integral becomes: $$\int \frac{3}{(\sin \theta - 3 \cos \theta - 1)} d\theta = \int \frac{3}{\frac{2t + 2t^2 -4}{1+t^2}} \cdot \frac{2}{1+t^2} dt = \int \frac{3 (1+t^2)}{2t + 2t^2 -4} \cdot \frac{2}{1+t^2} dt = \int \frac{6}{2t^2 + 2t -4} dt.$$ 5. Simplify denominator: $$2t^2 + 2t -4 = 2(t^2 + t - 2) = 2(t + 2)(t - 1).$$ 6. Thus the integral is: $$\int \frac{6}{2 (t + 2)(t - 1)} dt = \int \frac{3}{(t + 2)(t - 1)} dt.$$ 7. Use partial fractions: $$\frac{3}{(t+2)(t-1)} = \frac{A}{t+2} + \frac{B}{t-1}$$ Multiply both sides by \((t+2)(t-1)\): $$3 = A(t-1) + B(t+2) = (A+B)t + (-A + 2B).$$ 8. Equate coefficients: For \(t\): \(A+B=0 \Rightarrow B = -A\) Constant term: \(-A + 2B =3\) Substitute \(B = -A\): \(-A + 2(-A) = -A - 2A = -3A = 3 \Rightarrow A = -1, B=1\). 9. So: $$\int \frac{3}{(t+2)(t-1)} dt = \int \left( \frac{-1}{t+2} + \frac{1}{t-1} \right) dt = - \ln|t+2| + \ln|t-1| + C = \ln \left| \frac{t-1}{t+2} \right| + C.$$ 10. Substitute \(t = \tan(\frac{\theta}{2})\) back: $$\int \frac{3}{\sin \theta - 3 \cos \theta - 1} d\theta = \ln \left| \frac{\tan\frac{\theta}{2} - 1}{\tan \frac{\theta}{2} + 2} \right| + C.$$ 11. Now for the second integral: $$\int \frac{7 \sin \theta - \cos \theta + 2}{\sin \theta - 3 \cos \theta - 1} d\theta.$$ 12. Use the same substitution: Numerator substitution: $$7 \sin \theta = 7 \cdot \frac{2t}{1+t^2} = \frac{14t}{1+t^2},$$ $$- \cos \theta = - \frac{1 - t^2}{1+t^2} = \frac{t^2 -1}{1+t^2},$$ $$+2 = \frac{2(1+t^2)}{1+t^2}.$$ Combine numerator: $$\frac{14t + t^2 -1 + 2 + 2t^2}{1+t^2} = \frac{14t + 3t^2 +1}{1+t^2}.$$ 13. Recall denominator term from step 3: $$\sin \theta - 3 \cos \theta -1 = \frac{2t + 2t^2 -4}{1+t^2}.$$ 14. So integral reduces to: $$\int \frac{14t + 3t^2 +1}{2t + 2t^2 -4} \cdot \frac{2}{1+t^2} dt.$$ 15. Multiply numerator and denominator: $$\int \frac{(14t + 3t^2 +1) 2}{2t + 2t^2 -4} dt = \int \frac{2 (3t^2 +14 t +1)}{2(t^2 + t - 2)} dt = \int \frac{3t^2 + 14 t +1}{t^2 + t -2} dt.$$ 16. Perform polynomial division since degree numerator (2) = degree denominator (2): Divide \(3t^2 +14 t +1\) by \(t^2 + t -2\). 17. Quotient: Leading terms divide as \(3t^2 / t^2 = 3\). Multiply denominator by 3: $$3(t^2 + t - 2) = 3t^2 + 3t -6.$$ 18. Subtract: $$(3t^2 +14 t +1) - (3t^2 +3 t -6) = 11 t + 7.$$ 19. So: $$\frac{3t^2 +14 t +1}{t^2 + t -2} = 3 + \frac{11 t + 7}{t^2 + t -2}.$$ 20. Rewrite integral: $$\int 3 dt + \int \frac{11 t + 7}{t^2 + t -2} dt = 3 t + \int \frac{11 t + 7}{(t + 2)(t - 1)} dt + C.$$ 21. Resolve the partial fraction for \(\frac{11 t + 7}{(t+2)(t-1)}\): $$\frac{11 t + 7}{(t + 2)(t - 1)} = \frac{A}{t+2} + \frac{B}{t-1}.$$ 22. Multiply both sides by denominator: $$11 t + 7 = A (t -1) + B (t + 2) = (A + B) t + (- A + 2B).$$ 23. Equate coefficients: $$A + B = 11,$$ $$- A + 2 B = 7.$$ 24. From first, \(B = 11 - A\). Substitute into second: $$- A + 2(11 - A) = 7 \Rightarrow - A + 22 - 2A = 7 \Rightarrow -3 A = 7 - 22 = -15 \Rightarrow A = 5.$$ 25. Then, \(B = 11 - 5 = 6\). 26. So integrate: $$\int \frac{11 t + 7}{(t + 2)(t - 1)} dt = \int \left( \frac{5}{t + 2} + \frac{6}{t - 1} \right) dt = 5 \ln |t + 2| + 6 \ln |t - 1| + C.$$ 27. Final answer for the second integral in terms of \(t\): $$3 t + 5 \ln |t + 2| + 6 \ln |t - 1| + C.$$ 28. Substitute back \(t = \tan \left( \frac{\theta}{2} \right)\): $$\int \frac{7 \sin \theta - \cos \theta + 2}{\sin \theta - 3 \cos \theta - 1} d\theta = 3 \tan \left( \frac{\theta}{2} \right) + 5 \ln \left| \tan \left( \frac{\theta}{2} \right) + 2 \right| + 6 \ln \left| \tan \left( \frac{\theta}{2} \right) - 1 \right| + C.$$ 29. Problem (c): Given \(f(x) = \frac{x^2}{2x - 1}\), find turning points of \(y = f(x)\) and \(y = \frac{1}{f(x)}\) and classify them. 30. Find the first derivative of \(f(x)\) using quotient rule: $$f(x) = \frac{x^2}{2x - 1},$$ $$f'(x) = \frac{(2x)(2x - 1) - x^2 (2)}{(2x - 1)^2} = \frac{4 x^2 - 2x - 2 x^2}{(2x - 1)^2} = \frac{2 x^2 - 2 x}{(2x - 1)^2} = \frac{2 x (x - 1)}{(2x - 1)^2}.$$ 31. Set \(f'(x) = 0\) for turning points: $$2 x (x - 1) = 0 \Rightarrow x = 0 \text{ or } x = 1.$$ 32. Calculate corresponding \(y\) values: \(f(0) = \frac{0}{-1} = 0\), \(f(1) = \frac{1}{2 - 1} = 1\). 33. Find second derivative \(f''(x)\) to classify: $$f'(x) = \frac{2 x (x - 1)}{(2x - 1)^2}.$$ Use quotient rule, let numerator \(u = 2 x (x - 1) = 2 x^2 - 2 x\), denominator \(v = (2x - 1)^2\). 34. Compute \(u' = 4 x - 2\), and \(v' = 2 (2x - 1)(2) = 8 (2x - 1)\). 35. Then: $$f''(x) = \frac{(u' v - u v')}{v^2} = \frac{(4x - 2)(2x - 1)^2 - (2x^2 - 2x) 8 (2x - 1)}{(2x - 1)^4}.$$ 36. Evaluate at \(x=0\): Numerator: $$(4(0) - 2)(2(0) - 1)^2 - (2(0)^2 - 2(0)) 8 (2(0) - 1) = (-2)(-1)^2 - (0) 8 (-1) = -2 - 0 = -2,$$ denominator positive for \(x=0\), so \(f''(0) < 0\) sign. Thus at \(x=0\) the turning point is a local maximum. 37. Evaluate at \(x=1\): Numerator: $$(4(1)-2)(2(1)-1)^2 - (2(1)^2 - 2(1)) 8 (2(1)-1) = (4 - 2)(2 - 1)^2 - (2 - 2) 8 (2 - 1) = 2 \cdot 1 - 0 = 2,$$ denominator positive, so \(f''(1) > 0\) sign. Thus at \(x=1\) the turning point is a local minimum. 38. Now consider \(g(x) = \frac{1}{f(x)} = \frac{2x - 1}{x^2} = \frac{2x}{x^2} - \frac{1}{x^2} = \frac{2}{x} - \frac{1}{x^2}.$ 39. Find derivative: $$g'(x) = -\frac{2}{x^2} + \frac{2}{x^3} = -\frac{2}{x^2} + \frac{2}{x^3}.$$ Actually, apply quotient rule on \(g(x) = \frac{2x - 1}{x^2}\): 40. Using quotient rule: $$g'(x) = \frac{2 \cdot x^2 - (2x - 1) \cdot 2x}{x^4} = \frac{2 x^2 - 4 x^2 + 2 x}{x^4} = \frac{-2 x^2 + 2 x}{x^4} = \frac{2 x - 2 x^2}{x^4} = \frac{2 x (1 - x)}{x^4} = \frac{2 (1 - x)}{x^3}.$$ 41. Set \(g'(x) = 0\) for turning points: $$2 (1 - x) = 0 \Rightarrow x = 1.$$ 42. Compute second derivative: $$g''(x) = \frac{d}{dx} \left( \frac{2 (1 - x)}{x^3} \right) = 2 \frac{d}{dx} \left( (1 - x) x^{-3} \right).$$ 43. Use product rule: $$\frac{d}{dx} \left( (1 - x) x^{-3} \right) = -1 \cdot x^{-3} + (1 - x)(-3 x^{-4}) = - x^{-3} - 3 (1 - x) x^{-4} = - \frac{1}{x^3} - \frac{3 (1 - x)}{x^4}.$$ 44. Simplify common denominator \(x^4\): $$- \frac{x}{x^4} - \frac{3 (1 - x)}{x^4} = \frac{- x - 3 + 3 x}{x^4} = \frac{2 x - 3}{x^4}.$$ 45. So, $$g''(x) = 2 \cdot \frac{2 x - 3}{x^4} = \frac{4 x - 6}{x^4}.$$ 46. Evaluate at \(x=1\): $$g''(1) = \frac{4 (1) - 6}{1^4} = 4 - 6 = -2 < 0.$$ Thus, the turning point at \(x=1\) of \(g(x)\) is a local maximum. 47. Summarizing: - For \(f(x)\): \(x=0\): local maximum \((0,0)\) \(x=1\): local minimum \((1,1)\) - For \(g(x) = 1/f(x)\): \(x=1\): local maximum \(g(1) = \frac{1}{f(1)} = 1\) 48. The classifying test is based on second derivatives: \(f''(0) < 0\) max, \(f''(1) > 0\) min, \(g''(1) < 0\) max. Final answers provided with explanations and detailed steps.