Evaluate ( if convergent) $\int\limits_0^\infty \frac{1}{\sqrt{x(1-x)}} dx $ I would like to check whether the improper integral $$\int\limits_0^\infty \frac{1}{\sqrt{x(1-x)}} dx $$ is convergent or not. How can I check convergency? If convergent what is the integral value? Proceed: If we substitute $x=\sin^2\theta$, hten we have $$\int \frac1{\sqrt{x(1-x)}} \,dx = \sin^{-1} (2x-1) + C.$$ **Added:** for $x>1$, integrand is not defined. So if we take $0\leq x\leq 1$, then what we can say about its convergency?

The function $1/\sqrt{x(1-x)}$ is only defined over $(0,1)$, so the proposed integral makes no sense. If you want to compute $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx $$ your substitution is correct: set $x=\sin^2\theta$, with $\theta\in(0,\pi/2)$, so $$ dx=2\sin\theta\cos\theta\,d\theta $$ and you have $$ \int\frac{1}{\sqrt{x(1-x)}}\,dx= \int\frac{2\sin\theta\cos\theta}{\sqrt{\sin^2\theta\cos^2\theta}}\,d\theta= \int 2\,d\theta=2\theta+c $$ Thus your improper integral becomes $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx= \int_0^{\pi/2} 2\,d\theta=\Bigl[2\theta\Bigr]_0^{\pi/2}=\pi $$

Convergence follows also from the fact that $$ \int_0^1\frac{1}{\sqrt{x(1-x)}}\,dx= \int_0^{1/2}\frac{1}{\sqrt{x(1-x)}}\,dx \+ \int_{1/2}^1\frac{1}{\sqrt{x(1-x)}}\,dx = 2\int_0^{1/2}\frac{1}{\sqrt{x(1-x)}}\,dx $$ and this is convergent by comparison with $\int_0^{1/2}\frac{1}{\sqrt{x}}\,dx$