int(sqrt{1-sin x}+sqrt{1+sin x}) dx =f(x)+c, | vedclass

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(A) Given $I = \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx$.
Since $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1+\sin x = (\cos \frac{x}{2

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(A) Given $I = \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx$.
Since $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$, we have $\sqrt{1-\sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$ and $\sqrt{1+\sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$.
For $\frac{5 \pi}{2} < x < \frac{7 \pi}{2}$, we have $\frac{5 \pi}{4} < \frac{x}{2} < \frac{7 \pi}{4}$.
In this interval, $\cos \frac{x}{2} < 0$ and $\sin \frac{x}{2} < 0$, and specifically $\cos \frac{x}{2} < \sin \frac{x}{2}$.
Thus, $\sqrt{1-\sin x} = \sin \frac{x}{2} - \cos \frac{x}{2}$ and $\sqrt{1+\sin x} = -(\cos \frac{x}{2} + \sin \frac{x}{2})$.
$I = \int (\sin \frac{x}{2} - \cos \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2}) dx = \int -2 \cos \frac{x}{2} dx = -4 \sin \frac{x}{2} + C$.
So $f(x) = -4 \sin \frac{x}{2}$.
Then $f'(x) = -2 \cos \frac{x}{2}$.
$f'(\frac{8 \pi}{3}) = -2 \cos \frac{4 \pi}{3} = -2 (-\frac{1}{2}) = 1$.

$\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx =f(x)+c$, where $c$ is the constant of integration. If $\frac{5 \pi}{2}$

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