# BC Calculus: Review for Chapter 8 Worksheet

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

# Evaluate the following without a calculator:

1. $$\displaystyle \int_{0}^{2\sqrt{2}} \frac{x^3}{x^2 + 1} \, dx$$
2. Hint long divide, then u-substitute. Should lead to a ln.
Solution $$4 - \ln{3}$$
3. $$\displaystyle \int_{1}^{3} \frac{dy}{y^2 - 2y + 5}$$
4. Hint complete the square, u-sub, then use a reference triangle. Should lead to $$\tan^{-1}$$.
Solution $$\frac{\pi}{8}$$
5. $$\displaystyle \int_{\frac{1}{2}}^{1} \frac{\sqrt{1 - x^2}}{x^2} \, dx$$
6. Hint use a reference triangle. Should lead to $$\cot^2$$. Change to $$csc^2 - 1$$ (Pythagorean relationship) and integrate termwise.
Solution $$\sqrt{3} - \frac{\pi}{3}$$
7. $$\displaystyle \int_{1}^{4} \frac{dy}{y^2 - 2y + 10}$$
8. Hint complete the square, u-sub, and use a reference triangle. Should lead to an $$tan^{-1}$$.
Solution $$\frac{\pi}{12}$$
9. $$\displaystyle \int_{-2}^{2} \frac{x + 2}{\sqrt{x^2 + 4x + 13}} \, dx$$
10. Hint complete the square, u-sub, v-sub, then use the power rule.
Solution 2
11. $$\displaystyle \int_{0}^{5} \sqrt{25 - x^2} \, dx$$
12. Hint referemce triangle, then use the half angle formula. (Or, if you prefer to do it without calculus, it's just a quarter circle with radius = 5.)
Solution $$\frac{25 \pi}{4}$$
13. $$\displaystyle \int \frac{dx}{ \left( 4 - x^2 \right) ^{\frac{3}{2}}}$$
14. Hint use a reference triangle, reduce, and rewrite as secant sqaured. Then integrate.
Solution $$\frac{\frac{1}{4}x}{\sqrt{4 - x^2}} + C$$
15. $$\displaystyle \int \frac{dx}{\sqrt{x^2 + 2x}}$$
16. Hint complete the square, u-sub, then use a reference triangle. Should lead to integral of secant.
Solution $$\ln { \left| x + 1 + \sqrt{x^2 + 2x} \right| } + C$$
17. $$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2}{z\sqrt{4z^2 - 1}} \, dz$$
18. Hint use a reference triangle. Should lead to inverse secant.
Solution $$\frac{\pi}{3}$$
19. $$\displaystyle \int_{-1}^{-\frac{1}{2}} \frac{dx}{\sqrt{-2x - x^2}}$$
20. Hint complete the square, u-sub, then recognize it as the form of the inverse sine (or use a reference triangle.)
Solution $$\frac{\pi}{6}$$