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We start by using the trigonometric identity $sin~2x = 2~sin~x~cos~x$. The integral becomes: $$I = \int_{0}^{\frac{\pi}{2}} 2~sin~x~cos~x~tan^{-1}(sin~x)dx$$
Let $t = sin~x$. Then, $dt = cos~x~dx$. When $x = 0$, $t = sin~0 = 0$. When $x = \frac{\pi}{2}$, $t = sin~\frac{\pi}{2} = 1$. The integral transforms to: $$I = \int_{0}^{1} 2t~tan^{-1}(t)dt$$
We use integration by parts, where $u = tan^{-1}(t)$ and $dv = 2t~dt$. Then, $du = \frac{1}{1+t^2}dt$ and $v = t^2$. Using the formula $\int u~dv = uv - \int v~du$, we have: $$I = \left[t^2~tan^{-1}(t)\right]_{0}^{1} - \int_{0}^{1} \frac{t^2}{1+t^2}dt$$
Evaluate the first term: $$\left[t^2~tan^{-1}(t)\right]_{0}^{1} = (1^2~tan^{-1}(1)) - (0^2~tan^{-1}(0)) = 1 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4}$$
Simplify the second integral: $$\int_{0}^{1} \frac{t^2}{1+t^2}dt = \int_{0}^{1} \frac{1+t^2-1}{1+t^2}dt = \int_{0}^{1} \left(1 - \frac{1}{1+t^2}\right)dt$$
Evaluate the integral: $$\int_{0}^{1} \left(1 - \frac{1}{1+t^2}\right)dt = \left[t - tan^{-1}(t)\right]_{0}^{1} = (1 - tan^{-1}(1)) - (0 - tan^{-1}(0)) = 1 - \frac{\pi}{4}$$
Combine the results: $$I = \frac{\pi}{4} - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1$$
Final Answer: $\frac{\pi}{2} - 1$