For x in left(frac{3 pi}{4}, piright), evalua | vedclass

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(A) Given the integral $I = \int(\sqrt{1+\sin 2 x} + \sqrt{1-\sin 2 x}) \, dx$. Using the identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \

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$-2 \cos x + C$

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(A) Given the integral $I = \int(\sqrt{1+\sin 2 x} + \sqrt{1-\sin 2 x}) \, dx$. Using the identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$, we have: $\sqrt{1+\sin 2x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$ $\sqrt{1-\sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$ For $x \in \left(\frac{3 \pi}{4}, \pi\right)$, $\sin x > 0$ and $\cos x  |\cos x|$, so $\sin x + \cos x > 0$ and $\sin x - \cos x > 0$. Thus, the expression becomes $(\sin x + \cos x) + (\sin x - \cos x) = 2 \sin x$. Therefore, $I = \int 2 \sin x \, dx = -2 \cos x + C$.

For $x \in \left(\frac{3 \pi}{4}, \pi\right)$, evaluate the integral $\int(\sqrt{1+\sin 2 x}+\sqrt{1-\sin 2 x}) \, dx$.

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