int sqrt{1 + sin x} , dx = | vedclass

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Question

(D) We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2

Vedclass · Accepted Answer

$-2(\cos \frac{x}{2} - \sin \frac{x}{2}) + c$

Vedclass · Answer

(D) We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$. Thus,$\int \sqrt{1 + \sin x} \, dx = \int |\cos \frac{x}{2} + \sin \frac{x}{2}| \, dx$. Assuming the interval where $\cos \frac{x}{2} + \sin \frac{x}{2} > 0$,we have: $\int (\cos \frac{x}{2} + \sin \frac{x}{2}) \, dx = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + c = -2(\cos \frac{x}{2} - \sin \frac{x}{2}) + c$. Since $(\cos \frac{x}{2} - \sin \frac{x}{2})^2 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 - \sin x$,we can write the result as $-2\sqrt{1 - \sin x} + c$.

$\int \sqrt{1 + \sin x} \, dx = $

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