Integrate the function 1-x 1+x

Question

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

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Answer

Given integrand is; $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
Let $I= \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$
Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$
$\Rightarrow \sqrt{x}=\cos \theta \text { or } \theta=\cos ^{-1} \sqrt{x}$
$\Rightarrow \mathrm{I}=\int \sqrt{\frac{1-\sqrt{\cos ^{2} \theta}}{1+\sqrt{\cos ^{2} \theta}}}(-2 \sin \theta \cos \theta) \mathrm{d} \theta$
$=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta$
$=\int-\sqrt{\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}}(2 \sin \theta \cos \theta) d \theta$
= $\int-\sqrt{\frac{\sin ^{2}\left(\frac{\theta}{2}\right)}{\cos ^{2}\left(\frac{\theta}{2}\right)}}\left(2 \sin 2 \frac{\theta}{2} \cos 2 \frac{\theta}{2}\right) d \theta$
= $\int-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot(2) \cdot\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) \cdot\left(2 \cos ^{2}\left(\frac{\theta}{2}\right)-1\right) d \theta$
$\Rightarrow \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\int-4 \cdot\left[\sin ^{2}\left(\frac{\theta}{2}\right)\right]\left(2 \cos ^{2}\left(\frac{\theta}{2}\right)-1\right) d \theta$
$=\int-4 .\left\{\left[2 . \sin ^{2}\left(\frac{\theta}{2}\right) \cos ^{2}\left(\frac{\theta}{2}\right)\right]-\sin ^{2}\left(\frac{\theta}{2}\right)\right\} \mathrm{d} \theta$
$=\int-2 \cdot\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)^{2} d \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) d \theta$
$=-2 . \int \sin ^{2} \theta \mathrm{d} \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta$
$=-2 . \int \sin ^{2} \theta \mathrm{d} \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta$
= $-2 \cdot \int \frac{1-\cos 2 \theta}{2} d \theta+4 \int \frac{1-\cos \theta}{2} d \theta$
= $-2\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]+4\left[\frac{\theta}{2}-\frac{\sin \theta}{2}\right]+C$
= $-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C$
= $\theta+\frac{2 \cdot \sin \theta \cdot \cos \theta}{2}-2 \sin \theta+c$
= $\theta+\frac{2 \cdot \sqrt{1-\cos ^{2} \theta} \cdot \cos \theta}{2}-2 \sqrt{1-\cos ^{2} \theta}+C$
= $\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C$
= $\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}-2 \sqrt{1-x}+C$
$\Rightarrow \mathrm{I}=\cos ^{-1} \sqrt{\mathrm{x}}+\sqrt{\mathrm{x}-\mathrm{x}^{2}}-2 \sqrt{1-\mathrm{x}}+\mathrm{C}$

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