Question

\(\int \sqrt{1 + \csc x} dx\) is equal to

• A. \(\pm \sin^{-1}(\tan x - \sec x) + c\)
• B. \(2\sin^{-1}(\cos x) + c\)
• C. \(\sin^{-1}\left(\cos \frac{x}{2} - \sin \frac{x}{2}\right) + c\)
• D. \(\pm 2\sin^{-1}\left(\sin \frac{x}{2} - \cos \frac{x}{2}\right) + c\)

Hint:

Use half-angle formulas to simplify the integrand.

Answer 1

The Correct Option is D

Step 1: Formula / Definition}
\[ \sqrt{1 + \csc x} = \sqrt{\frac{\sin x + 1}{\sin x}} = \frac{\sin \frac{x}{2} + \cos \frac{x}{2}}{\sqrt{2\sin \frac{x}{2}\cos \frac{x}{2}}} \]
Step 2: Calculation / Simplification}
\(= \frac{\sin \frac{x}{2} + \cos \frac{x}{2}}{\sqrt{1 - (\sin \frac{x}{2} - \cos \frac{x}{2})^2}}\)
Let \(t = \sin \frac{x}{2} - \cos \frac{x}{2}\)
\(dt = \frac{1}{2}(\cos \frac{x}{2} + \sin \frac{x}{2})dx\)
\(\int \frac{2dt}{\sqrt{1-t^2}} = 2\sin^{-1}t + c = 2\sin^{-1}\left(\sin \frac{x}{2} - \cos \frac{x}{2}\right) + c\)
With \(\pm\) sign from square root.
Step 3: Final Answer
\[ \pm 2\sin^{-1}\left(\sin \frac{x}{2} - \cos \frac{x}{2}\right) + c \]