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If $\int e^{\sin ^2 x}\left(\sin x \cos x+\cos ^3 x \sin x\right) d x$ $=e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$
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The correct answer is:
$-\frac{1}{2} \sin 2 x$
$I=\int e^{\sin ^2 x}\left(\sin x \cos x+\cos ^3 x \sin x\right) d x$
$=\int e^{\sin ^2 x}\left(1+\cos ^2 x\right) \sin x \cos x d x$
$=\int e^{\sin ^2 x}\left(2-\sin ^2 x\right) \sin x \cos x d x$
Let $\sin ^2 x=t \Rightarrow \sin x \cos x d x=\frac{d t}{2}$
$\therefore I=\frac{1}{2} \int e^t(2-t) d t=e^t-\frac{1}{2}\left[t \cdot e^t-e^t\right]+C$
$=e^t\left[1-\frac{t}{2}+\frac{1}{2}\right]+C=e^{\sin ^2 x}\left[1+\frac{1}{2}-\frac{\sin ^2 x}{2}\right]+C$
$=e^{\sin ^2 x}[1+f(x)]+C$(given)
Therefore, $f(x)=\frac{1}{2}-\frac{\sin ^2 x}{2} \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sin 2 x$.
$=\int e^{\sin ^2 x}\left(1+\cos ^2 x\right) \sin x \cos x d x$
$=\int e^{\sin ^2 x}\left(2-\sin ^2 x\right) \sin x \cos x d x$
Let $\sin ^2 x=t \Rightarrow \sin x \cos x d x=\frac{d t}{2}$
$\therefore I=\frac{1}{2} \int e^t(2-t) d t=e^t-\frac{1}{2}\left[t \cdot e^t-e^t\right]+C$
$=e^t\left[1-\frac{t}{2}+\frac{1}{2}\right]+C=e^{\sin ^2 x}\left[1+\frac{1}{2}-\frac{\sin ^2 x}{2}\right]+C$
$=e^{\sin ^2 x}[1+f(x)]+C$(given)
Therefore, $f(x)=\frac{1}{2}-\frac{\sin ^2 x}{2} \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sin 2 x$.
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