Step-by-Step Solution
**Step 1: Define the integral**
Let $I = \int_0^{\pi/2} [\sin 2x \tan^{-1}(\sin x)] dx$
**Step 2: Use the identity $\sin 2x = 2 \sin x \cos x$**
$I = \int_0^{\pi/2} [2 \sin x \cos x \tan^{-1}(\sin x)] dx$
**Step 3: Use substitution**
Let $t = \sin x$, then $dt = \cos x dx$
When $x = 0$, $t = \sin 0 = 0$
When $x = \pi/2$, $t = \sin (\pi/2) = 1$
So, $I = \int_0^1 [2t \tan^{-1}(t)] dt$
**Step 4: Use integration by parts**
Let $u = \tan^{-1}(t)$ and $dv = 2t dt$
Then $du = \frac{1}{1+t^2} dt$ and $v = t^2$
Using integration by parts formula: $\int u dv = uv - \int v du$
$I = [t^2 \tan^{-1}(t)]_0^1 - \int_0^1 \frac{t^2}{1+t^2} dt$
**Step 5: Evaluate the first term and simplify the integral**
$I = (1^2 \tan^{-1}(1) - 0^2 \tan^{-1}(0)) - \int_0^1 \frac{t^2+1-1}{1+t^2} dt$
$I = \tan^{-1}(1) - \int_0^1 (1 - \frac{1}{1+t^2}) dt$
$I = \frac{\pi}{4} - \int_0^1 1 dt + \int_0^1 \frac{1}{1+t^2} dt$
**Step 6: Evaluate the integrals**
$I = \frac{\pi}{4} - [t]_0^1 + [\tan^{-1}(t)]_0^1$
$I = \frac{\pi}{4} - (1-0) + (\tan^{-1}(1) - \tan^{-1}(0))$
$I = \frac{\pi}{4} - 1 + \frac{\pi}{4} - 0$
$I = \frac{\pi}{2} - 1$