If frac{3 pi}{2}

Question

(B) We have $I = \int(\sqrt{1-\sin x} + \sqrt{1+\sin x}) \, dx$. Using $1 \pm \sin x = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2}\right)^2$, we get:

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) We have $I = \int(\sqrt{1-\sin x} + \sqrt{1+\sin x}) \, dx$. Using $1 \pm \sin x = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2}\right)^2$, we get: $I = \int \left( \sqrt{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} + \sqrt{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} \right) \, dx = \int \left( |\cos \frac{x}{2} - \sin \frac{x}{2}| + |\cos \frac{x}{2} + \sin \frac{x}{2}| \right) \, dx$. Given $\frac{3 \pi}{2} In this interval, $\sin \frac{x}{2} > \cos \frac{x}{2}$ and $\sin \frac{x}{2} + \cos \frac{x}{2} Thus, $|\cos \frac{x}{2} - \sin \frac{x}{2}| = \sin \frac{x}{2} - \cos \frac{x}{2}$ and $|\cos \frac{x}{2} + \sin \frac{x}{2}| = -(\cos \frac{x}{2} + \sin \frac{x}{2})$. $I = \int (\sin \frac{x}{2} - \cos \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2}) \, dx = \int -2 \cos \frac{x}{2} \, dx = -4 \sin \frac{x}{2} + c$. So, $f(x) = -4 \sin \frac{x}{2}$. Then $f\left(\frac{\pi}{3}\right) - f(0) = -4 \sin \frac{\pi}{6} - (-4 \sin 0) = -4 \left(\frac{1}{2}\right) + 0 = -2$.

If $\frac{3 \pi}{2} < x < \frac{5 \pi}{2}$ and $\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) \, dx = f(x) + c$ where $c$ is the constant of integration, then $f\left(\frac{\pi}{3}\right) - f(0) =$

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