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 \[\int \sqrt{2-x^2}\;d x=\left( \frac{x}{\sqrt{2}} \right) \sqrt{ 1-\frac{x^2}{2}} +\arcsin \left(\frac{x}{\sqrt{2}}\right) + C\]

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