If we let x = 4 sin θ {\textstyle x=4\sin \theta } , then d x = 4 cos θ d θ {\textstyle dx=4\cos \theta \,d\theta } and using the trigonometric identity sin 2 θ + cos 2 θ = 1 {\textstyle \sin ^{2}\theta +\cos ^{2}\theta =1} , we have 16 − x 2 = 16 ( 1 − sin 2 θ ) = 4 cos θ {\textstyle {\sqrt {16-x^{2}}}={\sqrt {16(1-\sin ^{2}\theta )}}=4\cos \theta } . Thus we get
Answer: π / 6. {\displaystyle \color {blue}\pi /6.}
![{\displaystyle \int _{0}^{2}{\frac {dx}{\sqrt {16-x^{2}}}}=\int _{\arcsin(0)}^{\arcsin(1/2)}{\frac {4\cos \theta \,d\theta }{4\cos \theta }}=\int _{0}^{\pi /6}d\theta ={\bigg [}\theta {\bigg ]}_{0}^{\pi /6}=\pi /6.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/92d540b640829e4a0d246ed96f4e1ecb953975bc)
