# 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

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