If I(x) = int e^{sin^2 x} (cos x sin 2x - sin | vedclass

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(C) Given $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. Using $\sin 2x = 2 \sin x \cos x$,we have: $I(x) = \int e^{\sin^2 x} (2 \sin x \cos

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$\frac{1}{2} e^{\frac{3}{4}}$

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(C) Given $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. Using $\sin 2x = 2 \sin x \cos x$,we have: $I(x) = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) dx$. Let $u = \sin x$,then $du = \cos x dx$. $I(x) = \int e^{u^2} (2u \cos x - 1) dx$. Alternatively,observe that $\frac{d}{dx} (e^{\sin^2 x} \cos x) = e^{\sin^2 x} (2 \sin x \cos x) \cos x + e^{\sin^2 x} (-\sin x) = e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) = e^{\sin^2 x} (\cos x \sin 2x - \sin x)$. Thus,$I(x) = e^{\sin^2 x} \cos x + C$. Given $I(0) = 1$,we have $1 = e^{\sin^2 0} \cos 0 + C \Rightarrow 1 = 1 \cdot 1 + C \Rightarrow C = 0$. So,$I(x) = e^{\sin^2 x} \cos x$. Then $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\pi/3)} \cos(\pi/3) = e^{3/4} \cdot \frac{1}{2} = \frac{1}{2} e^{\frac{3}{4}}$.

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

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