12 HL: Definite Integrals.

Complete as many of the following questions as you can before our next class (we’ll continue to work on these in our next class as well).

Exercise 22A questions 3–8, 12 (just complete the centre column for 3–8 and 12), 14, 16, 21
Exercise 22B questions 1adf, 2ad, 3, 4a

12 HL: Trigonometric Substitutions

In class we arrived at part of the solution to \(\int \sqrt{2-x^2}\;d x\), using the substitution \(x = \sqrt{2}\sin \theta\). Complete this question by showing that \[\int \sqrt{2-x^2}\;d x=\left( \frac{x}{\sqrt{2}} \right) \sqrt{ 1-\frac{x^2}{2}} +\arcsin \left(\frac{x}{\sqrt{2}}\right) + C\]

Hint
Note that, given our substitution choice, it follows that \(\sin \theta = \frac{x}{\sqrt{2}}\). Can we do something similar for \(\cos \theta\)?

Also complete

Exercise 21E questions 4, 5d, 6
Exercise 21F questions 12ad, 17, 18bd