If $$Ix = \int e^{\sin^2 x}  \sin 2x \cdot \sin^2 x \, dx$$ and $$I(0) = 1$$, then $$I\left(\frac{\pi}{3}\right)$$ is equal to

Let

$$t=\sin^2x.$$

Then,

$$dt=\sin2x,dx.$$

Therefore,

$$I(x)=\int te^t,dt.$$

Using integration by parts,

$$\int te^t,dt=e^t(t-1)+C.$$

Hence,

$$I(x)=e^{\sin^2x}\left(\sin^2x-1\right)+C.$$

Using

$$I(0)=1,$$

we get

$$e^0(0-1)+C=1$$

$$-1+C=1$$

$$C=2.$$

Therefore,

$$I(x)=e^{\sin^2x}\left(\sin^2x-1\right)+2.$$

Now,

$$\sin^2\frac{\pi}{3}=\frac34.$$

Hence,

$$I\left(\frac{\pi}{3}\right)=e^{3/4}\left(\frac34-1\right)+2$$

$$=2-\frac{e^{3/4}}{4}.$$

Therefore,

$$\boxed{2-\frac{e^{3/4}}{4}}$$