How do you find the integral of #int 1/(sqrtxsqrt(1-x)#?

Question

1161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-22T12:07:46

#2arcsin(sqrtx)+C#

#I=int1/(sqrtxsqrt(1-(sqrtx)^2))dx#

Let #u=sqrtx# so #du=1/(2sqrtx)dx#.

#I=2int1/(2sqrtxsqrt(1-(sqrtx)^2))dx#

#I=2int1/sqrt(1-u^2)du#

You may recognize this as the arcsine integral, but we can substitute #u=sintheta# to show this. We also see that #du=costhetad theta#.

#I=2int1/sqrt(1-sin^2theta)(costhetad theta)#

Since #1-sin^2theta=cos^2theta#:

#I=2int1/costhetacosthetad theta#

#I=2intd theta#

#I=2theta+C#

From #u=sintheta# we see that #theta=arcsin(u)#:

#I=2arcsin(u)+C#

#I=2arcsin(sqrtx)+C#