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

Given,

$$I(x)=\int e^{\sin^2 x}\cos x\sin 2x\cdot\sin x\,dx$$

Using

$$\sin 2x=2\sin x\cos x$$

the integrand becomes

$$e^{\sin^2 x}\cos x\cdot 2\sin x\cos x\cdot\sin x$$

$$=2e^{\sin^2 x}\sin^2 x\cos^2 x$$

Let

$$t=\sin^2 x$$

Then

$$dt=2\sin x\cos x\,dx$$

Rewrite the integrand as

$$e^{\sin^2 x}\sin x\cos x\,(2\sin x\cos x\,dx)$$

Therefore,

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

Integrating by parts,

$$\int te^t\,dt=t e^t-\int e^t\,dt$$

$$=te^t-e^t+C$$

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

Substituting

$$t=\sin^2 x$$

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

Using

$$I(0)=1$$

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

$$1=-1+C$$

$$C=2$$

Hence,

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

Now,

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

Therefore,

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

$$=2-\frac14e^{3/4}$$

Final Answer :

$$2-\frac14e^{3/4}$$