Question

\( \int_{0}^{\pi} \frac{1}{1+\sin x} \, dx \) is equal to

• A. 1
• B. 2
• C. -1
• D. -2

Hint:

Rationalizing the denominator $(\frac{1-\sin x}{1-\sin^2 x})$ also works quickly for this integral.

Answer 1

The Correct Option is B

Step 1: Trigonometric Identity
Using $\sin x = \frac{2 \tan(x/2)}{1 + \tan^{2}(x/2)}$. Integral becomes $\int_{0}^{\pi} \frac{\sec^{2}(x/2)}{(1 + \tan(x/2))^{2}} dx$.
Step 2: Substitution
Put $\tan(x/2) = t \implies \frac{1}{2} \sec^{2}(x/2) dx = dt$. Limits change from $[0, \pi]$ to $[0, \infty]$.
Step 3: Integration
$I = \int_{0}^{\infty} \frac{2 dt}{(1+t)^{2}} = \left[-\frac{2}{1+t}\right]_{0}^{\infty} = 0 - (-2) = 2$.
Final Answer: (b)