Find the integral $\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$ The question is $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$$ 
I have tried to multiply both numerator and denominator by $1-\sqrt{x}$ but can't proceed any further, help!

I think its useful to learn some standard substitution you can use for such kind of problem. 
for this case instead of using $x = \cos^2\theta$ I'm gonna try using $x=\cos^22\theta$ $$x=\cos^22\theta \Rightarrow dx=\left(-4\cos 2\theta \sin 2\theta \right)d\theta$$
so we have $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx$$
$$-4\int\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\cos 2\theta \left(\sin 2\theta \right),d\theta$$
$$-4\int \frac{\ \sin \theta }{\cos \theta }\cos 2\theta \left(\sin 2\theta \right)$$
$$-8\int (\sin^2 \theta\cos2\theta)d\theta $$
$$-8\int \frac{\ 1-\cos 2\theta }{2}\times \ (\cos 2\theta)d\theta $$
$$-4\int \cos 2\theta \ -\frac{\left[\cos 4\theta +1\right]}{2}d\theta  $$
$$-2\sin 2\theta +\frac{\sin 4\theta }{2}+(2\theta) + C$$
we want to write this in terms of $x$ so $\ \theta =\frac{\cos ^{-1}\left(\sqrt{\ x}\right)}{2}$
