Question

If \[ \int \frac{2}{1+\sin x} dx = 2 \log |A(x) - B(x)| + C \] and \( 0 \leq x \leq \frac{\pi}{2} \), then \( B(\pi/4) = \) ? 

• A. \( \frac{1}{\sqrt{2} + 3\sqrt{3}} \)
• B. \( \frac{1}{\sqrt{3} + 2\sqrt{2}} \)
• C. \( \frac{-1}{\sqrt{3} + 2\sqrt{2}} \)
• D. \( \frac{2}{\sqrt{2} + \sqrt{2}} \)

Hint:

For integrals involving trigonometric functions, use substitutions that simplify the denominator. Additionally, logarithmic transformations often help in obtaining closed-form expressions.

Answer 1

The Correct Option is B

Step 1: Solve the Integral 
We solve: \[ I = \int \frac{2}{1+\sin x} dx. \] Using the standard substitution: \[ I = \int \frac{2(1-\sin x)}{(1+\sin x)(1-\sin x)} dx. \] \[ = \int \frac{2(1-\sin x)}{\cos^2 x} dx. \] \[ = \int \frac{2}{\cos^2 x} dx - \int \frac{2\sin x}{\cos^2 x} dx. \] Since, \[ \int \sec^2 x dx = \tan x \quad \text{and} \quad \int \frac{\sin x}{\cos^2 x} dx = -\frac{1}{\cos x}. \] We get, \[ I = 2 \tan x + 2 \sec x + C. \] Thus, \[ I = 2 \log |A(x) - B(x)| + C. \] 
Step 2: Find \( B(\pi/4) \) 
Substituting \( x = \frac{\pi}{4} \): \[ B\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{3} + 2\sqrt{2}}. \] 
Step 3: Conclusion 
Thus, the value of \( B(\pi/4) \) is: \[ \boxed{\frac{1}{\sqrt{3} + 2\sqrt{2}}}. \]