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# Show that

Question:

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

Solution:

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

Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$

\begin{aligned} I &=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta \\ &=-\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \sin 2 \theta d \theta} \end{aligned}

$=-\int \tan \frac{\theta}{2} \cdot 2 \sin \theta \cos \theta d \theta$

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

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

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

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

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

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

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

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

$=-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+\mathrm{C}$

$=\theta+\frac{\sin 2 \theta}{2}-2 \sin \theta+\mathrm{C}$

$=\theta+\frac{2 \sin \theta \cos \theta}{2}-2 \sin \theta+\mathrm{C}$

$=\theta+\sqrt{1-\cos ^{2} \theta} \cdot \cos \theta-2 \sqrt{1-\cos ^{2} \theta}+\mathrm{C}$

$=\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+\mathrm{C}$

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

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