How do you evaluate #int dx/sqrt(4-x^2)# from [0,1]?

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Answer 1 · 2017-01-03T21:57:12

#pi/6#

Use the substitution #x = 2sintheta#. Then #dx = 2costheta d theta#.

#=>int_0^1 1/sqrt(4 - (2sintheta)^2) * 2costheta d theta#

#=>int_0^1 1/sqrt(4(1 - sin^2theta)) 2costheta d theta#

Apply the pythagorean identity #cos^2x = 1 - sin^2x#:

#=>int_0^1 1/sqrt(4cos^2theta) 2costheta d theta#

#=>int_0^1 1/(2costheta) 2costheta d theta#

#=>int_0^1 1 d theta#

#=> int_0^1 arcsin(x/2)#

Evaluate using the second fundamental theorem of calculus.

#=>arcsin(1/2) - arcsin(0/2)#

#=> pi/6#

Hopefully this helps!