Question

Prove that: \[ \int_0^{\pi/2} \sin 2x \tan^{-1} (\sin x) \,dx = \left(\frac{\pi}{2} - 1\right) \]

Hint:

Use the transformation property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) to simplify definite integrals.

Answer 1

Step 1: Use integral transformation property. Let \[ I = \int_0^{\pi/2} \sin 2x \tan^{-1} (\sin x) \,dx \] Using the property: \[ \int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx \] Substituting \( x = \frac{\pi}{2} - t \), we get: \[ I = \int_0^{\pi/2} \sin 2x \tan^{-1} (\cos x) \,dx \] Step 2: Add both integrals. \[ 2I = \int_0^{\pi/2} \sin 2x \left[ \tan^{-1} (\sin x) + \tan^{-1} (\cos x) \right] \,dx \] Using the identity: \[ \tan^{-1} (\sin x) + \tan^{-1} (\cos x) = \frac{\pi}{4} \] we get: \[ 2I = \frac{\pi}{4} \int_0^{\pi/2} \sin 2x \,dx \] Step 3: Compute the integral. \[ \int_0^{\pi/2} \sin 2x \,dx = \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2} \] \[ = \frac{1}{2} (1 - (-1)) = 1 \] \[ 2I = \frac{\pi}{4} \times 1 = \frac{\pi}{4} \] \[ I = \frac{\pi}{2} - 1 \] Final Answer: \[ I = \frac{\pi}{2} - 1 \]