Evaluation of Trigonometric Integration $\int \sin x\sqrt{\sin 2x}dx$. 
Evaluation of $$\int \sin x\sqrt{\sin 2x}dx$$

$\bf{My\; Try::}$ Let $$I = \int \sin x\sqrt{\sin 2x}dx = \sqrt{2}\int \frac{\tan x\sqrt{\tan x}}{\sec x}dx = \sqrt{2}\int \frac{(\tan x)^{\frac{3}{2}}}{\sec^3 x}\cdot \sec^2 xdx$$
So $$I = \sqrt{2}\int \frac{(\tan x)^{\frac{3}{2}}}{\sqrt{(1+\tan^2 x)^3}}\cdot \sec^2 xdx$$
Now Put $\tan x= t^2\;,$ Then $\sec^2 xdx = 2tdt$ 
So $$I = 2\sqrt{2}\int \frac{t^4}{\sqrt{(1+t^8)^3}}dt$$
Now how can i solve it after that, Help required, Thanks

Note that $\sin{2x} \equiv 1-(\sin{x}-\cos{x})^2 \equiv (\sin{x}+\cos{x})^2 -1$
Also note that $\text{d}(\sin{x}-\cos{x}) - \text{d}(\sin{x}+\cos{x}) = 2\sin{x}\,\text{d}x$
Substitute and separate the integral into two different integrals which can be evaluated independently.
