Question

Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Answer 1

$I=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Put $x={\cos }^2\theta$

$\Rightarrow dx=-2\sin \theta \cos \theta d\theta$

$I=\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}(-2\sin \theta \cos \theta )d\theta$

$=-2\int \sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}\left(\sin \theta \cos \theta \right)d\theta$

$=-2\int \frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\left(2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right)\cos \theta d\theta$

$=-4\int {\sin }^2\frac{\theta }{2}\cos \theta d\theta$

$=-4\int {\sin }^2\frac{\theta }{2}\cdot (2{\cos }^2\frac{\theta }{2}-1)d\theta$

$=-4\int (2{\sin }^2\frac{\theta }{2}{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2})d\theta$

$=-8\int {\sin }^2\frac{\theta }{2}\cdot {\cos }^2\frac{\theta }{2}d\theta +4\int {\sin }^2\frac{\theta }{2}d\theta$

$=-2\int {\sin }^2\theta d\theta +4\int {\sin }^2\frac{\theta }{2}d\theta$

$=-2\int \left(\frac{1-\cos 2\theta }{2}\right)d\theta +4\int \frac{1-\cos \theta }{2}d\theta$

$=-2[\frac{\theta }{2}-\frac{\sin 2\theta }{4}]+4[\frac{\theta }{2}-\frac{\sin \theta }{2}]+C$

$=-\theta +\frac{\sin 2\theta }{2}+2\theta -2\sin \theta +C$
$=\theta +\frac{\sin 2\theta }{2}-2\sin \theta +C$
$=\theta +\frac{2\sin \theta \cos \theta }{2}-2\sin \theta +C$
$=\theta +\sqrt{1-{\cos }^2\theta }\cdot \cos \theta -2\sqrt{1-{\cos }^2\theta }+C$
$={\cos }^{-1}\sqrt{x}+\sqrt{1-x}\cdot \sqrt{x}-2\sqrt{1-x}+C$
$=-2\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+\sqrt{x(1-x)}+C$
$=-2\sqrt{1-x}+co{s}^{-1}\sqrt{x}+\sqrt{x-x^2}+C$