Question

\[\int \sec^{- 1} \sqrt{x}\ dx\]

Answer 1

\[\int 1_{II} . \sec^{- 1} \sqrt{x}_I dx\]
\[ = \sec^{- 1} \sqrt{x}_{} \int1\text{  dx }- \int\left\{ \frac{d}{dx}\left( \sec^{- 1} \sqrt{x} \right)\int1 \text{ dx }\right\}dx\]
\[ = \sec^{- 1} \sqrt{x} . x - \int \frac{1}{\sqrt{x} \sqrt{1 - x}} \times \frac{1}{2\sqrt{x}} \times \text{  x dx }\]
\[ = x \sec^{- 1} \sqrt{x} - \frac{1}{2} \int \left( 1 - x \right)^{- \frac{1}{2}} dx\]
\[ = x \sec^{- 1} x - \frac{1}{2} \left[ \frac{\left( 1 - x \right)^{- \frac{1}{2} + 1}}{\left( - \frac{1}{2} + 1 \right) \left( - 1 \right)} \right] + C\]
\[ = x \sec^{- 1} x + \left( 1 - x \right)^\frac{1}{2} + C\]