Question

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

Answer 1

\[\text{We have}, \]

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

\[\text{ Putting } \sqrt{x} = \sec \theta\]

\[ \Rightarrow x = \sec^2 \theta\]

\[ \Rightarrow dx = 2 \text{ sec }\text{ θ } \text{ sec} \text{ θ  } \text{ tan   θ } \text{ dθ }\]

\[ = 2 \sec^2 \theta \text{ tan   θ } \text{ dθ }\]

\[ \therefore I = 2\int\theta \sec^2 \theta \text{ tan   θ } \text{ dθ }\]

\[ = 2 \int \theta\tan \theta \sec^2 \text{    θ } \text{ dθ }\]

\[\text{Considering}\text{  θ  as first fucction and} \tan \theta \sec^2 \ \text{theta as second function}\]

\[I = 2\left[ \theta\frac{\tan^2 \theta}{2} - \int1\frac{\tan^2 \theta}{2}d\theta \right]................ \left( \because \int\tan \theta \sec^2 \text{ tan   θ } \text{ dθ } = \frac{\tan^2 \theta}{2} \right)\]

\[ = \theta \tan^2 \theta - \int\left( \sec^2 \theta - 1 \right)d\theta\]

\[ = \theta \tan^2 \theta - \tan \theta + \theta + C\]

\[ = \theta\left( 1 + \tan^2 \theta \right) - \tan \theta + C\]

\[ = \theta \sec^2 \theta - \sqrt{se c^2 \theta - 1} + C\]

\[ = \sec^{- 1} \sqrt{x} x - \sqrt{x - 1} + C\]

\[ = x \sec^{- 1} \sqrt{x} - \sqrt{x - 1} + C\]