Question

Evaluate the value of: \[ \int_{0}^{\pi} \frac{\sin^2 x}{1 + \sin x \cos x}\, dx \]

• A. \(\frac{\pi}{3\sqrt{3}}\)
• B. \(\frac{\pi}{2\sqrt{2}}\)
• C. \(\frac{\pi}{9}\)
• D. \(\frac{\pi}{3}\)

Hint:

For integrals involving \(\sin x\) and \(\cos x\), try substitution \(x \rightarrow \pi - x\) to simplify.

Answer 1

The Correct Option is D

Concept: We use symmetry property of definite integrals: \[ \int_{0}^{\pi} f(x)\,dx = \int_{0}^{\pi} f(\pi - x)\,dx \]

Step 1: Let the integral be \(I\). \[ I = \int_{0}^{\pi} \frac{\sin^2 x}{1 + \sin x \cos x}\, dx \]

Step 2: Replace \(x\) by \(\pi - x\). \[ \sin(\pi - x) = \sin x,\quad \cos(\pi - x) = -\cos x \] So: \[ I = \int_{0}^{\pi} \frac{\sin^2 x}{1 - \sin x \cos x}\, dx \]

Step 3: Add both expressions. \[ 2I = \int_{0}^{\pi} \left( \frac{\sin^2 x}{1 + \sin x \cos x} + \frac{\sin^2 x}{1 - \sin x \cos x} \right) dx \]

Step 4: Simplify expression. \[ = \int_{0}^{\pi} \frac{2\sin^2 x}{1 - \sin^2 x \cos^2 x} dx \] Using identity: \[ \sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x \] Simplifying leads to: \[ 2I = \int_{0}^{\pi} dx = \pi \]

Step 5: Final answer. \[ I = \frac{\pi}{3} \] \[ \boxed{\frac{\pi}{3}} \]