## Integration ogf Reciprocal of Square Root of 1-x^n

To integrate
$I= \int^1_0 \frac{1}{ \sqrt{1-x^n}} dx$
substitute
$1-x^n=cos^2 \theta$
then
$-nx^{n-1} dx=-2sin \theta cos \theta d \theta$
.
Hence
\begin{aligned} dx &= \frac{2sin \theta cos \theta}{nx^{n-1}} d \theta = \frac{2xsin \theta cos \theta}{nx^n} d \theta = \frac{2(sin \theta )^{2/n} sin \theta cos \theta}{n(1-cos^2 \theta} d \theta \\ &= \frac{2(sin \theta )^{2/n} sin \theta cos \theta}{n(sin^2 \theta} d \theta =\frac{2sin^{(2-n)/n} cos \theta}{n} d \theta \end{aligned}
.
The integral becomes
\begin{aligned}I &= \int^1_0 \frac{1}{ \sqrt{1-x^n}} dx = \frac{2}{n} \int^{\pi /2}_0 \frac{sin^{(2-n)/n} cos \theta}{cos \Theta} d \theta \\ & = \int^{\pi /2}_0 sin^{(2-n)/n} d \theta \end{aligned}
.
From the beta function,
$\int^{\pi /2}_0 sin^n \theta d \theta = \int^{\pi /2}_0 cos^n \theta d \theta = \frac{\Gamma (\frac{n+1}{2}}{\Gamma (\frac{n}{2}+1} \frac{\sqrt{ \pi}}{2}$
.
Hence
$\int^1_0 \frac{1}{ \sqrt{1-x^n}} dx = \frac{2}{n} \int^{\pi /2}_0 sin^{(2-n)/n} d \theta = \frac{\Gamma (\frac{1}{n}}{\Gamma (\frac{1}{n}+ \frac{1}{2}} \frac{\sqrt{ \pi}}{2}$
.