^{-1}(x/2) evaluated between 1 and 2 which is an angle whose sine is 1-an angle whose sine is 1/2 which is π/2-π/6=π/3

Find ∫ 1/(sqrt(4-x^2)) dx from 1 to 2.

Answer: (1/3)pi

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the integral is sin^{-1}(x/2) evaluated between 1 and 2 which is an angle whose sine is 1-an angle whose sine is 1/2 which is π/2-π/6=π/3

Regards

Jim

∫[4 - x^{2}]^{-½} dx = sin^{-1} (x/2) evaluated between 1 and 2

sin^{-1}(2) - sin^{-1}(1/2) = (1/3)pi

William S., how did you get sin^-1 (x/2)?

The way you get arcsin(x/2)from ∫dx/√(4-x^{2})=∫dx/(2√(1-(x/2)^{2})=∫d(x/2)/(√(1-(x/2)^{2})

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## Comments

_{b→2}_{-}( ∫_{1}^{b}[4 - x^{2}]^{-½}dx )