Question

The value of the integral \(\int_{0}^{\pi} \frac{\cos x}{1 + \sin^{2} x} \, dx\) is

• A. 0
• B. 1
• C. \(\frac{\pi}{2}\)
• D. \(\pi\)
• E. \(2\pi\)

Hint:

Graphically, \(\cos x\) is antisymmetric about \(x = \pi/2\), while \(1 + \sin^2 x\) is symmetric. The integral of an "odd-like" symmetric function over such an interval often results in zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem can be solved using the properties of definite integrals. Specifically, the property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\) or by recognizing the symmetry of the function over the interval \([0, \pi]\).

Step 2: Key Formula or Approach:
We use the property: \[ I = \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]

Step 3: Detailed Explanation:
1. Let \( I = \int_{0}^{\pi} \frac{\cos x}{1 + \sin^{2} x} \, dx \).
2. Applying the property \(x \to \pi - x\): \[ I = \int_{0}^{\pi} \frac{\cos(\pi - x)}{1 + \sin^{2}(\pi - x)} \, dx \]
3. Using trigonometric identities \(\cos(\pi - x) = -\cos x\) and \(\sin(\pi - x) = \sin x\): \[ I = \int_{0}^{\pi} \frac{-\cos x}{1 + \sin^{2} x} \, dx = -I \]
4. Adding the equations: \(I + I = 0 \implies 2I = 0 \implies I = 0\).

Step 4: Final Answer
The value of the integral is 0.