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!

We have
$\mathrm{\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\ dx}$
Put $\mathrm{\sqrt{x} = \cos t}$
Differentiate Both Sides w.r.t. $\mathrm{x}$
$\mathrm{\frac{dx}{2\sqrt{x}}= - sint \ dt}$
$\mathrm{dx =- 2 \ cost\ sint \ dt}$
So, we get now:
$\mathrm{-\int \sqrt{\frac{1- cost}{1+ cost}}\  \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int \sqrt{\frac{2 \ sin^2 \frac{t}{2}}{2 \ cos^2 \frac{t}{2}}}\  \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int tan\frac{t}{2}  \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int \frac{sin\frac{t}{2}}{cos\frac{t}{2}}  \times 2\ \left( 2cos^2\frac{t}{2} -1    \right) \ sint \ dt}$
=  $\mathrm{-\int 4 \ sin\frac{t}{2}cos\frac{t}{2}sint \ dt + 2\int \frac{sin\frac{t}{2}}{cos\frac{t}{2}} sint \ dt}$
= $\mathrm{-2 \int sin^2 t dt + 4 \int sin^2 \frac{t}{2}dt}$
= $\mathrm{-2 \int \frac{1}{2}\ dt + 2 \int cos2t\ dt + 4 \int \frac{1}{2}dt - 4\int cost \ dt}$
=$\mathrm{-t + sin2t + 2t - 4 \ sint}$
= $\mathrm{t + sin2t - 4 \ sint}$
= $\mathrm{cos^{-1}\sqrt{x}+ 2\sqrt{x} \ sin(cos^{-1}\sqrt{x}) - 4 sin(cos^{-1}\sqrt{x})}$
= $\mathrm{cos^{-1}\sqrt{x} + 2\sqrt{x}(\sqrt{1-x}) - 4(\sqrt{1-x}) }$

= $\mathrm{cos^{-1}\sqrt{x} + 2(\sqrt{x(1-x)}) - 4(\sqrt{1-x}) + C}$

And, This is your final answer!
