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:

If replacing \(x\) by \(a-x\) changes sign of integrand, the integral over symmetric limits is zero.

Answer 1

The Correct Option is A

Concept: Use substitution and symmetry of definite integrals.

Step 1: Substitute \( t = \sin x \). Then: \[ dt = \cos x \, dx \] Limits change as: \[ x = 0 \Rightarrow t = 0, x = \pi \Rightarrow t = 0 \] So the integral becomes: \[ \int_{0}^{\pi} \frac{\cos x}{1+\sin^2 x} dx = \int_{0}^{0} \frac{dt}{1+t^2} \]

Step 2: Interpret limits carefully. Although limits appear same, we must note: As \(x\) goes from \(0 \to \pi\), \[ \sin x: 0 \to 1 \to 0 \] So split integral: \[ \int_0^\pi = \int_0^{\pi/2} + \int_{\pi/2}^\pi \]

Step 3: Use symmetry. Let: \[ I = \int_0^\pi \frac{\cos x}{1+\sin^2 x} dx \] Substitute \(x \to \pi - x\): \[ \cos(\pi - x) = -\cos x, \sin(\pi - x) = \sin x \] So: \[ I = \int_0^\pi \frac{-\cos x}{1+\sin^2 x} dx = -I \]

Step 4: Solve. \[ I = -I \Rightarrow 2I = 0 \Rightarrow I = 0 \]