How do you integrate #int 1/(sqrt[x]sqrt[4-x])#?

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Answer 1 · 2018-03-20T09:36:45

The answer is #=2arcsin(sqrtx/2)+C#

Perform the substitution

#sinu=sqrtx/2#, #=>#, #cosudu=1/(4sqrtx)dx#

Therefore,

The integral is

#I=int(dx)/(sqrtxsqrt(4-x))=int(4sqrtxcosudu)/(sqrtxsqrt(4-4sin^2u))#

#=2int1*du#

#=2u#

#=2arcsin(sqrtx/2)+C#