If x in left( frac{pi}{4}, frac{3pi}{4} right | vedclass

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(A) Given the integral $I = \int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx$. We know that $1 - \sin 2x = \sin^2 x + \cos^2 x

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$e^{\sin x} + c$

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(A) Given the integral $I = \int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx$. We know that $1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2$. Thus,$\sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$. Since $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$,we have $\sin x > \cos x$,so $|\sin x - \cos x| = \sin x - \cos x$. Substituting this into the integral: $I = \int \frac{\sin x - \cos x}{\sin x - \cos x} e^{\sin x} \cos x \, dx = \int e^{\sin x} \cos x \, dx$. Let $u = \sin x$,then $du = \cos x \, dx$. $I = \int e^u \, du = e^u + c = e^{\sin x} + c$.

If $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$,then $\int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx = $

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