1. Problem 25: Evaluate $$\int_0^{\pi/4} (1 + e^{\tan \theta}) \sec^2 \theta \, d\theta$$ using substitution. 2. Use substitution: let $$u = \tan \theta$$, then $$\frac{du}{d\theta} = \sec^2 \theta$$, so $$d\theta = \frac{du}{\sec^2 \theta}$$. 3. Substitute into the integral: $$\int_0^{\pi/4} (1 + e^{\tan \theta}) \sec^2 \theta \, d\theta = \int_0^1 (1 + e^u) \cancel{\sec^2 \theta} \frac{du}{\cancel{\sec^2 \theta}} = \int_0^1 (1 + e^u) du$$ 4. Integrate: $$\int_0^1 (1 + e^u) du = \int_0^1 1 \, du + \int_0^1 e^u \, du = [u]_0^1 + [e^u]_0^1 = (1 - 0) + (e - 1) = e$$ 5. Final answer for problem 25: $$e$$ 6. Problem 26: Evaluate $$\int_{\pi/4}^{\pi/2} (1 + e^{\cot \theta}) \csc^2 \theta \, d\theta$$ using substitution. 7. Let $$v = \cot \theta$$, then $$\frac{dv}{d\theta} = -\csc^2 \theta$$, so $$d\theta = \frac{dv}{-\csc^2 \theta}$$. 8. Substitute: $$\int_{\pi/4}^{\pi/2} (1 + e^{\cot \theta}) \csc^2 \theta \, d\theta = \int_1^0 (1 + e^v) \cancel{\csc^2 \theta} \frac{dv}{-\cancel{\csc^2 \theta}} = -\int_1^0 (1 + e^v) dv = \int_0^1 (1 + e^v) dv$$ 9. Integrate: $$\int_0^1 (1 + e^v) dv = [v]_0^1 + [e^v]_0^1 = 1 + (e - 1) = e$$ 10. Final answer for problem 26: $$e$$ 11. Problem 27: Evaluate $$\int_0^{\pi} \frac{\sin t}{2 - \cos t} dt$$ using substitution. 12. Let $$w = 2 - \cos t$$, then $$\frac{dw}{dt} = \sin t$$, so $$dw = \sin t \, dt$$. 13. Change limits: when $$t=0$$, $$w=2 - \cos 0 = 2 - 1 = 1$$; when $$t=\pi$$, $$w=2 - \cos \pi = 2 - (-1) = 3$$. 14. Substitute: $$\int_0^{\pi} \frac{\sin t}{2 - \cos t} dt = \int_1^3 \frac{1}{w} dw = [\ln |w|]_1^3 = \ln 3 - \ln 1 = \ln 3$$ 15. Final answer for problem 27: $$\ln 3$$ 16. Problem 28: Evaluate $$\int_0^{\pi} \frac{4 \sin \theta}{1 - 4 \cos \theta} d\theta$$ using substitution. 17. Let $$z = 1 - 4 \cos \theta$$, then $$\frac{dz}{d\theta} = 4 \sin \theta$$, so $$dz = 4 \sin \theta \, d\theta$$. 18. Change limits: when $$\theta=0$$, $$z=1 - 4 \cos 0 = 1 - 4 = -3$$; when $$\theta=\pi$$, $$z=1 - 4 \cos \pi = 1 - 4(-1) = 1 + 4 = 5$$. 19. Substitute: $$\int_0^{\pi} \frac{4 \sin \theta}{1 - 4 \cos \theta} d\theta = \int_{-3}^5 \frac{1}{z} dz = [\ln |z|]_{-3}^5 = \ln 5 - \ln 3 = \ln \frac{5}{3}$$ 20. Final answer for problem 28: $$\ln \frac{5}{3}$$ 21. Problem 29: Evaluate $$\int_1^2 \frac{2 \ln x}{x} dx$$ using substitution. 22. Let $$u = \ln x$$, then $$\frac{du}{dx} = \frac{1}{x}$$, so $$du = \frac{1}{x} dx$$. 23. Change limits: when $$x=1$$, $$u=0$$; when $$x=2$$, $$u=\ln 2$$. 24. Substitute: $$\int_1^2 \frac{2 \ln x}{x} dx = \int_0^{\ln 2} 2u \, du = 2 \int_0^{\ln 2} u \, du = 2 \left[ \frac{u^2}{2} \right]_0^{\ln 2} = (\ln 2)^2$$ 25. Final answer for problem 29: $$(\ln 2)^2$$ 26. Problem 30: Evaluate $$\int_2^4 \frac{dx}{x \ln x}$$ using substitution. 27. Let $$v = \ln x$$, then $$dv = \frac{1}{x} dx$$. 28. Change limits: when $$x=2$$, $$v=\ln 2$$; when $$x=4$$, $$v=\ln 4 = 2 \ln 2$$. 29. Substitute: $$\int_2^4 \frac{dx}{x \ln x} = \int_{\ln 2}^{2 \ln 2} \frac{1}{v} dv = [\ln |v|]_{\ln 2}^{2 \ln 2} = \ln (2 \ln 2) - \ln (\ln 2) = \ln 2$$ 30. Final answer for problem 30: $$\ln 2$$ 31. Problem 31: Evaluate $$\int_0^4 \frac{dx}{x (\ln x)^2}$$ using substitution. 32. Let $$w = \ln x$$, then $$dw = \frac{1}{x} dx$$. 33. Change limits: when $$x=0$$, $$w \to -\infty$$ (improper integral); when $$x=4$$, $$w=\ln 4$$. 34. Substitute: $$\int_0^4 \frac{dx}{x (\ln x)^2} = \int_{-\infty}^{\ln 4} \frac{1}{w^2} dw = \lim_{a \to -\infty} \int_a^{\ln 4} w^{-2} dw = \lim_{a \to -\infty} \left[-\frac{1}{w} \right]_a^{\ln 4} = \lim_{a \to -\infty} \left(-\frac{1}{\ln 4} + \frac{1}{a} \right) = -\frac{1}{\ln 4}$$ 35. Final answer for problem 31: $$-\frac{1}{\ln 4}$$ 36. Problem 32: Evaluate $$\int_2^{16} \frac{dx}{2x \sqrt{\ln x}}$$ using substitution. 37. Let $$u = \sqrt{\ln x}$$, so $$u^2 = \ln x$$, then $$2u du = \frac{1}{x} dx$$, so $$dx = 2x u du$$. 38. Substitute into integral: $$\int_2^{16} \frac{dx}{2x \sqrt{\ln x}} = \int_{u(2)}^{u(16)} \frac{2x u du}{2x u} = \int_{\sqrt{\ln 2}}^{\sqrt{\ln 16}} du = \sqrt{\ln 16} - \sqrt{\ln 2}$$ 39. Since $$\ln 16 = \ln (2^4) = 4 \ln 2$$, so $$\sqrt{\ln 16} = 2 \sqrt{\ln 2}$$. 40. Final answer for problem 32: $$2 \sqrt{\ln 2} - \sqrt{\ln 2} = \sqrt{\ln 2}$$ 41. Problem 33: Evaluate $$\int_0^{\pi/2} \frac{\tan x}{2} dx$$. 42. Rewrite integral: $$\int_0^{\pi/2} \frac{\tan x}{2} dx = \frac{1}{2} \int_0^{\pi/2} \tan x \, dx$$ 43. Integral of $$\tan x$$ is $$-\ln |\cos x|$$. 44. Evaluate: $$\frac{1}{2} [-\ln |\cos x|]_0^{\pi/2} = \frac{1}{2} (-\ln 0 + \ln 1)$$ which diverges to infinity. 45. Final answer for problem 33: The integral diverges (does not converge). 46. Problem 34: Evaluate $$\int_{\pi/4}^{\pi/2} \cot t \, dt$$. 47. Integral of $$\cot t$$ is $$\ln |\sin t|$$. 48. Evaluate: $$[\ln |\sin t|]_{\pi/4}^{\pi/2} = \ln 1 - \ln \frac{\sqrt{2}}{2} = -\ln \frac{\sqrt{2}}{2} = \ln \sqrt{2}$$ 49. Final answer for problem 34: $$\ln \sqrt{2}$$ 50. Problem 35: Evaluate $$\int_0^{\pi/3} \tan^2 \theta \cos \theta \, d\theta$$. 51. Rewrite $$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$$, so integrand is $$\frac{\sin^2 \theta}{\cos^2 \theta} \cos \theta = \frac{\sin^2 \theta}{\cos \theta}$$. 52. Let $$u = \sin \theta$$, then $$du = \cos \theta d\theta$$. 53. Substitute: $$\int_0^{\pi/3} \tan^2 \theta \cos \theta d\theta = \int_0^{\sin \pi/3} \frac{u^2}{\cos \theta} \frac{du}{\cos \theta}$$ but this is complicated; better to use substitution directly. 54. Alternatively, rewrite integral as: $$\int_0^{\pi/3} \tan^2 \theta \cos \theta d\theta = \int_0^{\pi/3} \tan^2 \theta \cos \theta d\theta$$ 55. Use substitution $$v = \sin \theta$$, then $$dv = \cos \theta d\theta$$. 56. Then: $$\int_0^{\pi/3} \tan^2 \theta \cos \theta d\theta = \int_0^{\sin \pi/3} \left(\frac{v}{\cos \theta}\right)^2 dv$$ but $$\cos \theta = \sqrt{1 - v^2}$$. 57. So integrand becomes: $$\frac{v^2}{1 - v^2} dv$$. 58. Integral: $$\int_0^{\sqrt{3}/2} \frac{v^2}{1 - v^2} dv = \int_0^{\sqrt{3}/2} \frac{v^2 - 1 + 1}{1 - v^2} dv = \int_0^{\sqrt{3}/2} \left(-1 + \frac{1}{1 - v^2}\right) dv$$ 59. Split integral: $$\int_0^{\sqrt{3}/2} -1 dv + \int_0^{\sqrt{3}/2} \frac{1}{1 - v^2} dv = -\frac{\sqrt{3}}{2} + \int_0^{\sqrt{3}/2} \frac{1}{1 - v^2} dv$$ 60. Integral of $$\frac{1}{1 - v^2}$$ is $$\frac{1}{2} \ln \left| \frac{1+v}{1-v} \right|$$. 61. Evaluate: $$-\frac{\sqrt{3}}{2} + \left[ \frac{1}{2} \ln \left| \frac{1+v}{1-v} \right| \right]_0^{\sqrt{3}/2} = -\frac{\sqrt{3}}{2} + \frac{1}{2} \ln \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}$$ 62. Final answer for problem 35: $$-\frac{\sqrt{3}}{2} + \frac{1}{2} \ln \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}$$ 63. Problem 36: Evaluate $$\int_0^{\pi/12} 6 \tan 3x \, dx$$. 64. Let $$u = 3x$$, then $$du = 3 dx$$, so $$dx = \frac{du}{3}$$. 65. Change limits: when $$x=0$$, $$u=0$$; when $$x=\pi/12$$, $$u=\pi/4$$. 66. Substitute: $$\int_0^{\pi/12} 6 \tan 3x \, dx = \int_0^{\pi/4} 6 \tan u \frac{du}{3} = 2 \int_0^{\pi/4} \tan u \, du$$ 67. Integral of $$\tan u$$ is $$-\ln |\cos u|$$. 68. Evaluate: $$2 [-\ln |\cos u|]_0^{\pi/4} = 2 (-\ln \frac{\sqrt{2}}{2} + \ln 1) = 2 \ln \sqrt{2} = \ln 2$$ 69. Final answer for problem 36: $$\ln 2$$