Question

Evaluate: \[ \int \tan^{-1}\left( \sqrt{\frac{1-\sin x}{1+\sin x}} \right)\,dx \]

• A. \(\dfrac{\pi x}{2}-\dfrac{x^2}{4}+C\)
• B. \(\dfrac{\pi x}{4}-\dfrac{x^2}{2}+C\)
• C. \(\dfrac{\pi x}{4}-\dfrac{x}{4}+C\)
• D. \(\dfrac{\pi x}{4}-\dfrac{x^2}{4}+C\)

Hint:

Remember the identity: \[ \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) = \sqrt{\frac{1-\sin x}{1+\sin x}} \] It appears frequently in integration problems involving inverse trigonometric functions.

Answer 1

The Correct Option is D

Concept: Complicated trigonometric expressions inside inverse trigonometric functions can often be simplified using standard identities. The important identity used here is: \[ \sqrt{\frac{1-\sin x}{1+\sin x}} = \frac{1-\sin x}{\cos x} = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \] This converts the integrand into a very simple form.

Step 1: Simplify the expression inside \(\tan^{-1}\). Consider: \[ \sqrt{\frac{1-\sin x}{1+\sin x}} \] Multiply numerator and denominator inside the root by \(1-\sin x\): \[ = \sqrt{ \frac{(1-\sin x)^2}{1-\sin^2x} } \] Using: \[ 1-\sin^2x=\cos^2x \] we get: \[ = \sqrt{ \frac{(1-\sin x)^2}{\cos^2x} } \] \[ = \frac{1-\sin x}{\cos x} \] Now use the standard identity: \[ \frac{1-\sin x}{\cos x} = \tan\left( \frac{\pi}{4}-\frac{x}{2} \right) \] Hence the integrand becomes: \[ \tan^{-1} \left[ \tan\left( \frac{\pi}{4}-\frac{x}{2} \right) \right] \] Therefore, \[ = \frac{\pi}{4}-\frac{x}{2} \]

Step 2: Integrate the simplified expression. Thus, \[ \int \tan^{-1}\left( \sqrt{\frac{1-\sin x}{1+\sin x}} \right)\,dx = \int \left( \frac{\pi}{4}-\frac{x}{2} \right)dx \] Integrating term-by-term: \[ = \frac{\pi x}{4} -\frac12\cdot\frac{x^2}{2} +C \] \[ = \frac{\pi x}{4} -\frac{x^2}{4} +C \] Hence, \[ \boxed{ \frac{\pi x}{4}-\frac{x^2}{4}+C } \]