Question

\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]

Answer 1

\[\text{ Let  I } = \int\frac{\sqrt{1 - x}}{\sqrt{x}}dx\]
\[ = \int\left( \frac{\sqrt{1 - x} \cdot \sqrt{1 - x}}{\sqrt{x} \cdot \sqrt{1 - x}} \right) dx\]
\[ = \int\frac{\left( 1 - x \right)}{\sqrt{x - x^2}}dx\]
\[\text{ Let} \left( 1 - x \right) = A\frac{d}{dx}\left( x - x^2 \right) + B\]
\[ \Rightarrow 1 - x = A \left( 1 - 2x \right) + B\]
\[ \Rightarrow 1 - x = - \left( 2A \right) x + A + B\]
\[\text{Equating coefficients of like terms}\]
\[ - 2A = - 1\]
\[ \Rightarrow A = \frac{1}{2}\]
\[\text{ and   A + B = 1 }\]
\[ \Rightarrow \frac{1}{2} + B = 1\]
\[ \therefore B = \frac{1}{2}\]
\[ \therefore I = \int\frac{\frac{1}{2} \left( 1 - 2x \right) + \frac{1}{2}}{\sqrt{x - x^2}}dx\]
\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{x - x^2 + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\]
\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x^2 - x + \frac{1}{2^2} \right)}}dx\]
\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}dx\]

 

 

\[\text{ Putting x - x}^2 =\text{  t in the first integral }\]

\[ \Rightarrow \left( 1 - 2x \right)\text{  dx } = dt\]

\[ \therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}dx\]

\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\]

\[ = \frac{1}{2} \times 2\text{  t}^\frac{1}{2} + \frac{1}{2} \times \sin^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right) + C................ \left[ \because \int\frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\]

\[ = \sqrt{t} + \frac{1}{2} \text{ sin}^{- 1} \left( 2x - 1 \right) + C\]

\[ = \sqrt{x - x^2} + \frac{1}{2} \text{ sin}^{- 1} \left( 2x - 1 \right) + C ..................\left[ \because t = x - x^2 \right]\]