Question

Evaluate: $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$

Answer 1

Put $x={\cos }^{2}2t⟹dx=-\sin 4tdt$

$\therefore I=\int \sqrt{\frac{1-\cos 2t}{1+\cos 2t}}-\sin 4tdt$

$=\int -\tan t\sin 4tdt$ (using $\cos 2t=co{s}^{2}t-si{n}^{2}t$ and $1=co{s}^{2}t+si{n}^{2}t$

$=\int -\tan t\sin 4tdt$

$=\int -\tan t2\sin 2t\cos 2tdt$ (using $\sin 2x=2\sin x\cos x$)

$=\int -\tan t4\sin t\cos t\cos 2tdt$ (using $\sin 2x=2\sin x\cos x$)

$=\int -4{\sin }^{2}t(2{\cos }^{2}t-1)dt$ (using $\cos 2x=2{\cos }^{2}x-1$)

$=\int (-8{\sin }^{2}t{\cos }^{2}t+4{\sin }^{2}t)dt$

$=\int (-2{\sin }^{2}2t+2(1-\cos 2t)dt$

$=\int (\cos 4t-1+2(1-\cos 2t)dt$

$=\frac{1}{4}\sin 4t-t+2t-\sin 2t$

$=\frac{1}{4}\sin 4t-t+2t-\sin 2t$

$=\frac{{\cos }^{-1}\sqrt{x}}{2}-\sqrt{1-x}+c$