Question

\( \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \) is equal to:

• A. \( \pi \)
• B. \( 0 \) (Zero)
• C. \( \int_{0}^{\pi/2} \frac{2 \sin x}{1 + \sin x \cos x} \, dx \)
• D. \( \frac{\pi}{4} \)

Hint:

For symmetric limits in definite integrals, substitution \( x \to \frac{\pi}{2} - x \) can simplify evaluation.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
The given integral is: \[ I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx. \] Let \( x \to \frac{\pi}{2} - x \). Then, \( \sin x \to \cos x \) and \( \cos x \to \sin x \). Substituting: \[ I = \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx. \] Step 2: Add and simplify.
Adding the original and transformed integrals: \[ 2I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx + \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx = 0. \] Hence: \[ I = 0. \]