P.o.t.W. #17 Solution
(a) \displaystyle\int {\sqrt {1 - {x^2}} } \,{\textrm{d}}x x = \sin {\theta }\;\;\; \Rightarrow \;\;\;{\textrm{d}}x = \cos {\theta }\,{\textrm{d\theta }} substituting: \displaystyle\int {\sqrt {1 - {x^2}} } \,{\textrm{d}}x = \int {\sqrt {1 - {{\sin }^2}{\theta }} } \cos {\theta }\,{\textrm{d\theta }} = \int {\cos {\theta }\cos {\theta }} \,{\textrm{d\theta }} = \int {{{\cos }^2}{\theta }} \,{\textrm{d\theta }}...