Question

Evaluate the integral: $ \int_0^{\frac{\pi}{2}} \frac{1}{1 + \sin x} \, dx $

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{\pi}{3} \)

Hint:

When faced with integrals involving trigonometric identities, consider multiplying the integrand by conjugates or using known reduction formulas.

Answer 1

The Correct Option is A

We are tasked with evaluating the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{1}{1 + \sin x} \, dx \] We can use the standard trigonometric identity to simplify the integrand. Multiply the numerator and denominator by \( 1 - \sin x \): \[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx \] Using the identity \( (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x \), the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \sin x}{\cos^2 x} \, dx \] Now, split the integrand: \[ I = \int_0^{\frac{\pi}{2}} \frac{1}{\cos^2 x} \, dx - \int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos^2 x} \, dx \] The first integral is \( \sec^2 x \), and its integral is \( \tan x \). 
The second integral is \( \frac{\sin x}{\cos^2 x} = \frac{d}{dx}(\tan x) \), so its integral is \( -\sec x \). 
Therefore: \[ I = \left[ \tan x \right]_0^{\frac{\pi}{2}} - \left[ \sec x \right]_0^{\frac{\pi}{2}} \] Evaluating: \[ I = \left( \tan \frac{\pi}{2} - \tan 0 \right) - \left( \sec \frac{\pi}{2} - \sec 0 \right) \] Since \( \tan \frac{\pi}{2} \to \infty \), this simplifies to: \[ I = \frac{\pi}{4} \]