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

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Answer 1 · 2015-09-27T01:47:26

#pi/6#

#int_1^2 1/(sqrtx sqrt(4-x))dx=int_1^2 dx/sqrt(4x-x^2)=#

#=int_1^2 dx/sqrt(4-4+4x-x^2)=int_1^2 dx/sqrt(4-(x^2-4x+4))=#

#=int_1^2 dx/sqrt(4-(x-2)^2)=1/2int_1^2 dx/sqrt(1-((x-2)/2)^2)=I#

#(x-2)/2=t => dx=2dt#

#x=1 => t=-1/2#

#x=2 => t=0#

#I=int_(-1/2)^0 dt/sqrt(1-t^2)=arcsint|_(-1/2)^0#

#I=0-(-pi/6)=pi/6#