Question

Evaluate \[ \int_{0}^{\pi/2} \frac{x\sin 2x}{1+4\cos^2 2x}\,dx. \]

• A. \[ \frac{\pi}{8}\tan^{-1}2 \]
• B. \[ \frac{\pi}{4}\tan^{-1}2 \]
• C. \[ \frac{\pi}{2}\tan^{-1}2 \]
• D. \[ \frac{\pi}{8}\tan^{-1}4 \]

Hint:

For definite integrals involving \(x\), always check whether \[ f(a-x)=f(x). \] If true, use \[ \int_0^a x f(x)\,dx = \frac{a}{2}\int_0^a f(x)\,dx, \] which greatly simplifies the calculation.

Answer 1

The Correct Option is A

Concept: Use the property \[ \int_0^a x\,f(x)\,dx = \frac{a}{2}\int_0^a f(x)\,dx \] whenever \[ f(a-x)=f(x). \]

Step 1: Verify the symmetry condition. Let \[ f(x) = \frac{\sin 2x}{1+4\cos^2 2x}. \] Then \[ f\!\left(\frac{\pi}{2}-x\right) = \frac{\sin(\pi-2x)} {1+4\cos^2(\pi-2x)}. \] Using \[ \sin(\pi-2x)=\sin2x, \] and \[ \cos^2(\pi-2x)=\cos^22x, \] we get \[ f\!\left(\frac{\pi}{2}-x\right) = f(x). \] Hence, \[ \int_0^{\pi/2}x\,f(x)\,dx = \frac{\pi}{4} \int_0^{\pi/2}f(x)\,dx. \] Therefore, \[ I = \frac{\pi}{4} \int_0^{\pi/2} \frac{\sin2x}{1+4\cos^22x}\,dx. \]

Step 2: Evaluate the remaining integral. Let \[ t=\cos2x. \] Then \[ dt=-2\sin2x\,dx, \] or \[ \sin2x\,dx=-\frac12\,dt. \] When \[ x=0, \quad t=1, \] and when \[ x=\frac{\pi}{2}, \quad t=-1. \] Thus, \[ \int_0^{\pi/2} \frac{\sin2x}{1+4\cos^22x}\,dx = \frac12 \int_{-1}^{1} \frac{dt}{1+4t^2}. \] \[ = \frac12 \left[ \frac12\tan^{-1}(2t) \right]_{-1}^{1}. \] \[ = \frac14 \Big( \tan^{-1}2-\tan^{-1}(-2) \Big). \] \[ = \frac12\tan^{-1}2. \]

Step 3: Find \(I\). \[ I = \frac{\pi}{4} \left( \frac12\tan^{-1}2 \right). \] \[ = \frac{\pi}{8}\tan^{-1}2. \] Therefore, \[ \boxed{ \int_{0}^{\pi/2} \frac{x\sin2x}{1+4\cos^22x}\,dx = \frac{\pi}{8}\tan^{-1}2 } \] \[ \boxed{\text{Answer = (A)}} \]