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Integrate the function
$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=I($ say $)$
$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=I($ say $)$
Solution:
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Verified Answer
Let $x=\cos ^2 t \Rightarrow d x=2 \cos t(-\sin t) d t$
$\begin{aligned}
&\therefore I=\int \sqrt{\frac{1-\cos t}{1+\cos t}}(2 \sin t \cos t) d t \\
&=-4 \int \frac{1-\cos t}{2} \cos t d t \\
&=-2 \int\left(\cos t-\frac{1+\cos 2 t}{2}\right) d t \\
&=-[2 \sin t-t-\sin t \cos t]+c \\
&=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x} \cdot \sqrt{1-x}+c
\end{aligned}$
$\begin{aligned}
&\therefore I=\int \sqrt{\frac{1-\cos t}{1+\cos t}}(2 \sin t \cos t) d t \\
&=-4 \int \frac{1-\cos t}{2} \cos t d t \\
&=-2 \int\left(\cos t-\frac{1+\cos 2 t}{2}\right) d t \\
&=-[2 \sin t-t-\sin t \cos t]+c \\
&=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x} \cdot \sqrt{1-x}+c
\end{aligned}$
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