1. **Problem a:** Find an appropriate trigonometric substitution for $$\int (5x^2 - 3)^{3/2} \, dx$$ given $$x = \sqrt{\frac{3}{5}} \sec \theta$$. Step 1: Recognize the form inside the integral: $$5x^2 - 3$$. Step 2: Substitute $$x = \sqrt{\frac{3}{5}} \sec \theta$$. Step 3: Then, $$5x^2 - 3 = 5 \left(\sqrt{\frac{3}{5}} \sec \theta\right)^2 - 3 = 5 \cdot \frac{3}{5} \sec^2 \theta - 3 = 3 \sec^2 \theta - 3 = 3(\sec^2 \theta - 1) = 3 \tan^2 \theta.$$ Step 4: So, $$(5x^2 - 3)^{3/2} = (3 \tan^2 \theta)^{3/2} = 3^{3/2} (\tan^2 \theta)^{3/2} = 3^{3/2} \tan^3 \theta.$$ Step 5: Also, compute $$dx$$: $$dx = \sqrt{\frac{3}{5}} \sec \theta \tan \theta \, d\theta.$$ Step 6: Substitute into the integral to simplify. --- 2. **Problem b:** Find an appropriate trigonometric substitution for $$\int \frac{x^2}{\sqrt{5x^2 + 4}} \, dx$$ given $$x = \frac{2}{\sqrt{5}} \tan \theta$$. Step 1: Substitute $$x = \frac{2}{\sqrt{5}} \tan \theta$$. Step 2: Then, $$5x^2 + 4 = 5 \left(\frac{2}{\sqrt{5}} \tan \theta\right)^2 + 4 = 5 \cdot \frac{4}{5} \tan^2 \theta + 4 = 4 \tan^2 \theta + 4 = 4(\tan^2 \theta + 1) = 4 \sec^2 \theta.$$ Step 3: So, $$\sqrt{5x^2 + 4} = 2 \sec \theta.$$ Step 4: Also, $$x^2 = \left(\frac{2}{\sqrt{5}} \tan \theta\right)^2 = \frac{4}{5} \tan^2 \theta,$$ and $$dx = \frac{2}{\sqrt{5}} \sec^2 \theta \, d\theta.$$ Step 5: Substitute into the integral to simplify. --- 3. **Problem c:** Find an appropriate trigonometric substitution for $$\int x \sqrt{3x^2 + 24x + 44} \, dx$$ given $$x = \frac{2}{\sqrt{3}} \sec \theta - 4$$. Step 1: Complete the square inside the root: $$3x^2 + 24x + 44 = 3(x^2 + 8x) + 44 = 3(x^2 + 8x + 16 - 16) + 44 = 3((x+4)^2 - 16) + 44 = 3(x+4)^2 - 48 + 44 = 3(x+4)^2 - 4.$$ Step 2: Substitute $$x + 4 = \frac{2}{\sqrt{3}} \sec \theta$$. Step 3: Then, $$3(x+4)^2 - 4 = 3 \left(\frac{2}{\sqrt{3}} \sec \theta\right)^2 - 4 = 3 \cdot \frac{4}{3} \sec^2 \theta - 4 = 4 \sec^2 \theta - 4 = 4(\sec^2 \theta - 1) = 4 \tan^2 \theta.$$ Step 4: So, $$\sqrt{3x^2 + 24x + 44} = 2 \tan \theta.$$ Step 5: Also, $$dx = \frac{2}{\sqrt{3}} \sec \theta \tan \theta \, d\theta.$$ Step 6: Substitute into the integral to simplify. --- 4. **Problem d:** Find an appropriate trigonometric substitution for $$\int \frac{x}{\sqrt{-66 - 8x^2 + 48x}} \, dx$$ given $$x = 3 + \frac{\sqrt{3}}{2} \sin \theta$$. Step 1: Complete the square inside the root: $$-66 - 8x^2 + 48x = -8x^2 + 48x - 66 = -8(x^2 - 6x) - 66 = -8(x^2 - 6x + 9 - 9) - 66 = -8((x-3)^2 - 9) - 66 = -8(x-3)^2 + 72 - 66 = -8(x-3)^2 + 6.$$ Step 2: Substitute $$x - 3 = \frac{\sqrt{3}}{2} \sin \theta$$. Step 3: Then, $$-8(x-3)^2 + 6 = -8 \left(\frac{\sqrt{3}}{2} \sin \theta\right)^2 + 6 = -8 \cdot \frac{3}{4} \sin^2 \theta + 6 = -6 \sin^2 \theta + 6 = 6(1 - \sin^2 \theta) = 6 \cos^2 \theta.$$ Step 4: So, $$\sqrt{-66 - 8x^2 + 48x} = \sqrt{6} \cos \theta.$$ Step 5: Also, $$dx = \frac{\sqrt{3}}{2} \cos \theta \, d\theta.$$ Step 6: Substitute into the integral to simplify. --- **Summary:** Each substitution transforms the integral into a trigonometric integral by rewriting the quadratic expressions inside radicals or powers as perfect squares involving trigonometric functions, simplifying the integration process.