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

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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 $theta=arcsin(u)$ :

$I=2arcsin(u)+C$

$I=2arcsin(sqrtx)+C$

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