How to calculate $\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} dx$? How to calculate?
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\, \mathrm dx$$
I try to let $x=\cos^2 t$, then
$$\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}=\tan\frac t2,\; dx=-2\sin t\cos t\,\mathrm dt $$
so
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} \mathrm dx=-2 \int\tan\frac t2\sin t\cos t\,\mathrm dt$$
Thanks a lot!

Alternate method:
$$ I = \int \sqrt{\frac{1- \sqrt{x}}{1 + \sqrt{x}}}\ dx$$
Putting $x = \cos^{2}{2t}$ so that $dx = -4\cos(2t) \sin(2t)\ dt$ and $t = \frac12 \cos^{-1}({\sqrt x})$
$$\begin{align}\implies I & = \int \frac{\sin(t)}{\cos(t)}\cdot -4\cos(2t) \sin(2t)\ dt\\ & = -4\int \frac{\sin t}{\cos t}[2\cos^2t - 1]\cdot 2 \sin t \cos t \ dt\\& = -8 \int\sin^2(t)[2\cos^2t - 1]\ dt\\& = -8 \int \frac{\sin^22t}{2} - \sin^2t \ dt\\& = -8 \int \frac{1 - \cos4t}{4} - \frac{1- \cos2t}{2} \ dt\\& = 2 \int \cos4t - 1\ dt + 4\int 1- \cos2t \ dt\\& = \frac{\sin 4t}{2} - 2t  + C_1 + 4t - 2\sin 2t+ C_2\\& = \sin(2t) \cos(2t) + 2t - 2\sin 2t+ C\\& = \sqrt{x - x^2} + \cos^{-1}\sqrt{x}- 2\sqrt{1-x} + C\end{align}$$
