Question

The integral $\displaystyle\int \frac{dx}{(1+ \sqrt{x}) \sqrt{x - x^2}}$ is equal to (where $C$ is a constant of integration)

• A. $-2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C$
• B. $-2 \sqrt{\frac{1- \sqrt{x}}{1+ \sqrt{x}}} + C$
• C. $- \sqrt{\frac{1- \sqrt{x}}{1+\sqrt{x}}} + C$
• D. $2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C$

Answer 1

The Correct Option is B

$I =\int \frac{ dx }{(1+\sqrt{ x }) \sqrt{ x - x ^{2}}}$
put $x =\cos ^{2} \theta$
$dx =-2 \cos \theta \sin \theta d \theta$
$I =\int \frac{-2 \sin \theta \cos \theta d \theta}{(1+\cos \theta) \cos \theta \sin \theta} \,\,\,\,\,=-2 \int \frac{ d \theta}{2 \cos ^{2} \theta / 2}$
$=-\int \sec ^{2}\left(\frac{\theta}{2}\right) d \theta \,\,\,\,\,\,\, \therefore \cos \theta=\sqrt{ x }$
$=-2 \tan \theta / 2+ C \,\,\,\,\,\, \frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}=\sqrt{ x }$
$=-2 \sqrt{\frac{1-\sqrt{ x }}{1+\sqrt{ x }}}+ c $
$\Rightarrow \tan ^{2}\left(\frac{\theta}{2}\right)=\frac{1-\sqrt{ x }}{1+\sqrt{ x }}$

Answer 2

The correct answer is B:\(-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\)
\(I=\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2)}}=\int\frac{dx}{(1+\sqrt{x})\sqrt{x}.(1-x)^{\frac{1}{2}}}\)
Let \(1+\sqrt{x}=p\)
\(\therefore\) \(\frac{dx}{2\sqrt{x}}=dp\)
\(\therefore I=\int \frac{2dp}{p\sqrt{2p-p^2}}\)
again put \(p=\frac{1}{t}\)
\(\therefore dp=\frac{-1}{t^2}dt\)
⇒\(I=2\int \frac{n\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{\frac{2}{t}-\frac{1}{t^2}}}\)
⇒\(I=-2\sqrt{2t-1}\)
\(\therefore -2\sqrt{\frac{2}{t}-1}+C\)
=\(-2\sqrt{\frac{2-t}{t}}\)
\(=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\)