Question

The value of \( \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\tan x + \sin x}{1 + \cos^2 x} \, dx \) is equal to:

• A. 0
• B. 2
• C. \( \sqrt{2} \)
• D. \( 2\sqrt{2} \)
• E. \( -2\sqrt{2} \)

Hint:

When integrating odd functions over symmetric intervals, the result is always zero: \[ \int_{-a}^{a} f(x) \, dx = 0. \] This is a fundamental property of integrals.

Answer 1

The Correct Option is A

We evaluate the integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\tan x + \sin x}{1 + \cos^2 x} \, dx. \] Step 1: Symmetry Property We check whether the function inside the integral is odd or even. Define: \[ f(x) = \frac{\tan x + \sin x}{1 + \cos^2 x}. \] Substituting \( x \to -x \): \[ f(-x) = \frac{\tan(-x) + \sin(-x)}{1 + \cos^2(-x)} = \frac{-\tan x - \sin x}{1 + \cos^2 x} = -f(x). \] Since \( f(-x) = -f(x) \), the function is odd. The integral of an odd function over a symmetric interval \( [-a, a] \) is always zero: \[ I = 0. \] 
Final Answer: (A) 0.