If int e^{sin ^2 x}(sin x cos x+cos ^3 x sin | vedclass

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(D) Let $I = \int e^{\sin ^2 x}(\sin x \cos x + \cos ^3 x \sin x) d x$. Factor out $\sin x \cos x$: $I = \int e^{\sin ^2 x}(1 + \cos ^2 x) \sin x \cos

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$-\frac{1}{2} \sin 2 x$

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(D) Let $I = \int e^{\sin ^2 x}(\sin x \cos x + \cos ^3 x \sin x) d x$. Factor out $\sin x \cos x$: $I = \int e^{\sin ^2 x}(1 + \cos ^2 x) \sin x \cos x d x$. Since $\cos ^2 x = 1 - \sin ^2 x$, we have: $I = \int e^{\sin ^2 x}(1 + 1 - \sin ^2 x) \sin x \cos x d x = \int e^{\sin ^2 x}(2 - \sin ^2 x) \sin x \cos x d x$. Let $t = \sin ^2 x$, then $dt = 2 \sin x \cos x d x$, so $\sin x \cos x d x = \frac{dt}{2}$. Substituting into the integral: $I = \frac{1}{2} \int e^t(2 - t) d t = \int e^t d t - \frac{1}{2} \int t e^t d t$. Using integration by parts for $\int t e^t d t = t e^t - e^t$: $I = e^t - \frac{1}{2}(t e^t - e^t) + C = e^t(1 - \frac{t}{2} + \frac{1}{2}) + C = e^t(\frac{3}{2} - \frac{t}{2}) + C$. Comparing with $e^{\sin ^2 x}(1 + f(x)) + C$, we have $1 + f(x) = \frac{3}{2} - \frac{\sin ^2 x}{2}$. Thus, $f(x) = \frac{1}{2} - \frac{\sin ^2 x}{2}$. Taking the derivative: $f^{\prime}(x) = 0 - \frac{1}{2}(2 \sin x \cos x) = -\sin x \cos x = -\frac{1}{2} \sin 2 x$.

If $\int e^{\sin ^2 x}(\sin x \cos x+\cos ^3 x \sin x) d x = e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

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