If I(x) = ∫ e ^ ^ 2 x ( x 2x - x)dx and I(0) = 1 , then I((π)/(3))...

Given,

$I\left(x\right) = ∫ {e}^{\sin^{2}x}\left(\cos x \sin 2x - \sin x\right)dx$

$⇒ I\left(x\right) = ∫ {e}^{\sin^{2}x}\left(\cos x - \dfrac{1}{2\cos x}\right)\sin 2xdx$

Now let, $\sin^{2}x = t ⇒ \sin 2xdx = dt$

$I\left(x\right) = ∫ {e}^{t}\left(\sqrt{1 - t} - \dfrac{1}{2\sqrt{1 - t}}\right)dt$

Now comparing with,

$I\left(x\right) = ∫ {e}^{t}\left(f\left(t\right) + {f}^{'}\left(t\right)\right)dt = {e}^{t}f\left(t\right) + c$

Here, $f\left(t\right) = \sqrt{1 - t} \text{ and } {f}^{'}\left(t\right) = - \dfrac{1}{2\sqrt{1 - t}}$

$⇒ I\left(x\right) = {e}^{t}\sqrt{1 - t} + c = {e}^{\sin^{2}x} \cdot \cos x + c$

Now using, $I\left(0\right) = 1 ⇒ c = 0$

Hence, the value of $I\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}e\dfrac{3}{4}$