Find ∫ 1/(sqrt(4-x^2)) dx from 1 to 2. Answer: (1/3)pi 11/10/2013 | Sun from Los Angeles, CA | 2 Answers | 0 Votes Mark favorite Subscribe Comment Comments Note that this is an improper integral since the integrand is undefined at 2. The integral is defined as lim_{b→2}_{-} ( ∫_{1}^{b} [4 - x^{2}]^{-½} dx ) 11/10/2013 | Andre W. Comment
Note that this is an improper integral since the integrand is undefined at 2. The integral is defined as lim_{b→2}_{-} ( ∫_{1}^{b} [4 - x^{2}]^{-½} dx ) 11/10/2013 | Andre W.
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 11/10/2013 | Jim S. Comment
∫[4 - x^{2}]^{-½} dx = sin^{-1} (x/2) evaluated between 1 and 2 sin^{-1}(2) - sin^{-1}(1/2) = (1/3)pi 11/10/2013 | William S. Comment Comments William S., how did you get sin^-1 (x/2)? 11/11/2013 | Sun from Los Angeles, CA 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}) 11/16/2013 | Michael F. Comment
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}) 11/16/2013 | Michael F.
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