show that $\int_0^1\frac{(dx)}{\sqrt{1-x^2}}=\frac{(1)}{(n)}*\beta({\frac{(1)}{(2)}},\frac{(1)}{(n)})$

Question

Show that $\int_0^1\frac{(dx)}{\sqrt{1-x^2}}=\frac{(1)}{(n)}*\beta({\frac{(1)}{(2)}},\frac{(1)}{(n)})$ ,when $n\in\mathbb{Z}^+$.

It was hard for me to prove it .

I integrated the left hand side and the result is $\frac{(\pi)}{(2)}$.

I thought that if I substitute the value of $n$ in the R.H.S I would get the same result as the L.H.S, but that didn't happen .

Is my thinking correct? How can I solve it ?

Answer 1

$$ \int_{0}^{1}\frac{1}{\sqrt{1-x^{2}}}dx=\left. \arcsin (x)\right\vert _{0}^{1}=\frac{\pi }{2} $$

How do you get the right side?

Answer 2

Substituting $x=\sin^{2/n}t$,$$\int_0^1x^{n/2-1}(1-x^n)^{1/n-1}dx=\int_0^{\pi/2}\frac2n\cos^{2/n-1}tdt=\frac1n\operatorname{B}\left(\frac12,\,\frac1n\right).$$Your problem is the special case $n=2$. If you found $n=2$ didn't work, you've miscalculated $\left.\frac1n\operatorname{B}\left(\frac12,\,\frac1n\right)\right|_{n=2}$.