What is #int 1/(sqrt[x]*sqrt[4-x]) dx#?

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Answer 1 · 2018-04-06T16:10:05

#2sin^(-1)(sqrtx/2)+c#

#I=int 1/(sqrt[x]*sqrt[4-x]) dx#?

substitute #u=sqrtx/2=>x=4u^2#

#=>du=1/4x^(-1/2)#

#:. dx=4sqrtxdu#

#I=int1/(cancel(sqrtx)(sqrt(4-4u^2)))4cancel(sqrtx)du#

#I=int4/(2sqrt(1-u^2))du#

#=2int(du)/(sqrt(1-u^2))#

this is astandard integral

#=2sin^(-1)u+c#

#=2sin^(-1)(sqrtx/2)+c#

Answer 2 · 2018-04-06T16:12:54

#=-2sin^-1(sqrt(4-x)/2)+C#

We know that,

#(I)color(red)(int1/sqrt(a^2-x^2)dx=sin^-1(x/a)+c#

Here,

#I=int1/(sqrtx*sqrt(4-x))dx#

Let, #sqrt(4-x)=u=>4-x=u^2#

i.e. #x=4-u^2=>dx=-2udu#

So,

#I=int(-2u)/(sqrt(4-u^2)*u)du#

#=-2int1/sqrt(2^2-u^2)du...toApply (I)#

#=-2sin^-1(u/2)+C,where,u=sqrt(4-x)#

#=-2sin^-1(sqrt(4-x)/2)+C#