1. Problem 16: Evaluate the integral $$\int x^2 e^{x^3} \, dx$$. 2. To solve this, use substitution. Let $$u = x^3$$, then $$\frac{du}{dx} = 3x^2$$ or $$du = 3x^2 dx$$. 3. Rearranging, $$x^2 dx = \frac{du}{3}$$. 4. Substitute into the integral: $$\int x^2 e^{x^3} dx = \int e^u \frac{du}{3} = \frac{1}{3} \int e^u du$$. 5. The integral of $$e^u$$ is $$e^u$$, so $$\frac{1}{3} e^u + C$$. 6. Substitute back $$u = x^3$$: $$\frac{1}{3} e^{x^3} + C$$. 7. Therefore, the answer to problem 16 is option c) $$\frac{1}{3} e^{x^3} + C$$. 8. Problem 17: Given $$y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right)$$, find $$\frac{dy}{dx}$$. 9. Let $$z = \frac{1 - \cos x}{\sin x}$$. Then $$y = \tan^{-1}(z)$$. 10. The derivative of $$\tan^{-1}(z)$$ with respect to $$x$$ is $$\frac{1}{1+z^2} \cdot \frac{dz}{dx}$$. 11. Compute $$\frac{dz}{dx}$$ using quotient rule: $$z = \frac{1 - \cos x}{\sin x}$$, Numerator $$u = 1 - \cos x$$, $$u' = \sin x$$. Denominator $$v = \sin x$$, $$v' = \cos x$$. $$\frac{dz}{dx} = \frac{u'v - uv'}{v^2} = \frac{\sin x \cdot \sin x - (1 - \cos x) \cdot \cos x}{\sin^2 x} = \frac{\sin^2 x - \cos x + \cos^2 x}{\sin^2 x}$$. 12. Use identity $$\sin^2 x + \cos^2 x = 1$$: $$\sin^2 x - \cos x + \cos^2 x = (\sin^2 x + \cos^2 x) - \cos x = 1 - \cos x$$. 13. So, $$\frac{dz}{dx} = \frac{1 - \cos x}{\sin^2 x}$$. 14. Next, compute $$1 + z^2$$: $$1 + \left( \frac{1 - \cos x}{\sin x} \right)^2 = 1 + \frac{(1 - \cos x)^2}{\sin^2 x} = \frac{\sin^2 x + (1 - \cos x)^2}{\sin^2 x}$$. 15. Expand numerator: $$(1 - \cos x)^2 = 1 - 2 \cos x + \cos^2 x$$. So numerator is: $$\sin^2 x + 1 - 2 \cos x + \cos^2 x = (\sin^2 x + \cos^2 x) + 1 - 2 \cos x = 1 + 1 - 2 \cos x = 2 - 2 \cos x = 2(1 - \cos x)$$. 16. Therefore, $$1 + z^2 = \frac{2(1 - \cos x)}{\sin^2 x}$$. 17. Now, $$\frac{dy}{dx} = \frac{1}{1 + z^2} \cdot \frac{dz}{dx} = \frac{1}{\frac{2(1 - \cos x)}{\sin^2 x}} \cdot \frac{1 - \cos x}{\sin^2 x} = \frac{\sin^2 x}{2(1 - \cos x)} \cdot \frac{1 - \cos x}{\sin^2 x} = \frac{1}{2}$$. 18. Hence, the derivative $$\frac{dy}{dx} = \frac{1}{2}$$. 19. The answer to problem 17 is option a) $$\frac{1}{2}$$.