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: Understanding the Concept:
We need to evaluate a definite integral of a trigonometric function. A common technique for integrals involving \( 1 \pm \sin x \) or \( 1 \pm \cos x \) in the denominator is to multiply the numerator and denominator by the conjugate.
Step 2: Key Formula or Approach:
1. Multiply the numerator and denominator by \( (1 - \sin x) \).
2. Use the Pythagorean identity \( 1 - \sin^2 x = \cos^2 x \).
3. Split the resulting fraction and integrate term by term.
Step 3: Detailed Explanation:
The integral is \( \int_0^{\pi/2} \frac{1}{1+\sin x} dx \).
Multiply the numerator and denominator by \( 1 - \sin x \):
\[ \int_0^{\pi/2} \frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} dx = \int_0^{\pi/2} \frac{1-\sin x}{1-\sin^2 x} dx \] Using the identity \( 1-\sin^2 x = \cos^2 x \):
\[ \int_0^{\pi/2} \frac{1-\sin x}{\cos^2 x} dx \] Split the integral into two parts:
\[ \int_0^{\pi/2} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx = \int_0^{\pi/2} (\sec^2 x - \tan x \sec x) dx \] Now, find the antiderivative:
\[ \int (\sec^2 x - \tan x \sec x) dx = \tan x - \sec x \] Evaluate the definite integral using the limits from 0 to \( \pi/2 \):
\[ [\tan x - \sec x]_0^{\pi/2} \] This presents a problem because \( \tan(\pi/2) \) and \( \sec(\pi/2) \) are undefined. This method, while standard for indefinite integrals, requires careful handling of limits for the definite integral. Alternative Method (Weierstrass Substitution):
Let \( t = \tan(x/2) \). Then \( dx = \frac{2 dt}{1+t^2} \) and \( \sin x = \frac{2t}{1+t^2} \).
The limits of integration change:
- When \( x = 0 \), \( t = \tan(0) = 0 \).
- When \( x = \pi/2 \), \( t = \tan(\pi/4) = 1 \).
The integral becomes:
\[ \int_0^1 \frac{1}{1 + \frac{2t}{1+t^2}} \cdot \frac{2 dt}{1+t^2} = \int_0^1 \frac{1}{\frac{1+t^2+2t}{1+t^2}} \cdot \frac{2 dt}{1+t^2} \] \[ = \int_0^1 \frac{1+t^2}{(t+1)^2} \cdot \frac{2 dt}{1+t^2} = \int_0^1 \frac{2}{(t+1)^2} dt \] Now integrate:
\[ \int 2(t+1)^{-2} dt = 2 \frac{(t+1)^{-1}}{-1} = \frac{-2}{t+1} \] Evaluate from 0 to 1:
\[ \left[ \frac{-2}{t+1} \right]_0^1 = \left( \frac{-2}{1+1} \right) - \left( \frac{-2}{0+1} \right) = \frac{-2}{2} - (-2) = -1 + 2 = 1 \] Step 4: Final Answer:
The value of the integral is 1.