Integrate the function in Exercise: 1- x 1+ x

Question

Integrate the function in Exercise:

$\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}$

✓

Answer

$\text{I}=\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}\text{dx}$
$\text{Let }\ \text{x}=\cos^{2}\theta\Rightarrow\text{dx}=-2\sin\theta\cos\theta\ \text{d}\theta$
$\text{I}=\int\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-2\sin\theta\cos\theta)\text{d}\theta$
$=-\int\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin2\theta\ \text{d}\theta$
$=-\int\tan\frac{\theta}{2}.\sin\theta\cos\theta\ \text{d}\theta$
$=-2\int\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\bigg(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\bigg)\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}.\bigg(2\cos^{2}\frac{\theta}{2}-1\bigg) \text{d}\theta$
$=-4\int\bigg(2\sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}\bigg) \text{d}\theta$
$=-8\int\sin^{2}\frac{\theta}{2}.\cos^{2}\frac{\theta}{2}\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\sin^{2}\theta\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\bigg(\frac{1-\cos2\theta}{2}\bigg)\text{d}\theta+4\int\frac{1-\cos\theta}{2}\text{d}\theta$
$=-2\bigg[\frac{\theta}{2}-\frac{\sin2\theta}{4}\bigg]+4\bigg[\frac{\theta}{2}-\frac{\sin\theta}{2}\bigg]+\text{C}$
$=-\theta+\frac{\sin2\theta}{2}+2\theta-2\sin\theta+\text{C}$
$=\theta+\frac{\sin2\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\frac{2\sin\theta\cos\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\sqrt{1-\cos^{2}\theta}.\cos\theta-2\sqrt{1-\cos^{2}\theta}+\text{C}$
$=\cos^{-1}\sqrt{\text{x}}+\sqrt{1-\text{x}}.\sqrt{\text{x}}-2\sqrt{1-\text{x}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}(1-\text{x)}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}-\text{x}^{2}}+\text{C}$