Question

$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin x}\,dx =$

• A. 2
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. 1
• E. 0

Hint:

The substitution $t = \tan(x/2)$ is a universal method for integrals involving $\sin x$ and $\cos x$.

Answer 1

The Correct Option is D

Step 1: Concept
Rationalize the integrand by multiplying the numerator and denominator by $(1 - \sin x)$.

Step 2: Analysis
$\int_{0}^{\pi/2} \frac{1-\sin x}{1-\sin^2 x} dx = \int_{0}^{\pi/2} \frac{1-\sin x}{\cos^2 x} dx = \int_{0}^{\pi/2} (\sec^2 x - \sec x \tan x) dx$.

Step 3: Calculation
$[\tan x - \sec x]_{0}^{\pi/2}$ is undefined at $\pi/2$, so we use the half-angle substitution or simplify: $\int_{0}^{\pi/2} \frac{1}{1+\sin x} dx = [\tan(x/2 - \pi/4)]$. The definite integral evaluates to 1. Final Answer: (D)