int sqrt{1 + sin x} , dx is equal to | vedclass

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(A) We know that $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$.Thus, $\sqrt{1 + \sin x} = |\cos(x/2) + \s

Vedclass · Accepted Answer

$2\sqrt{2} \sin \left( \frac{x}{2} - \frac{\pi}{4} \right) + c$

Vedclass · Answer

(A) We know that $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$.Thus, $\sqrt{1 + \sin x} = |\cos(x/2) + \sin(x/2)|$.Assuming the interval where $\cos(x/2) + \sin(x/2) > 0$, we have $\int (\cos(x/2) + \sin(x/2)) \, dx$.$= 2\sin(x/2) - 2\cos(x/2) + c$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we can write $2\sin(x/2) - 2\cos(x/2) = 2\sqrt{2} \left( \frac{1}{\sqrt{2}}\sin(x/2) - \frac{1}{\sqrt{2}}\cos(x/2) \right) = 2\sqrt{2} \sin(x/2 - \pi/4) + c$.

$\int \sqrt{1 + \sin x} \, dx$ is equal to

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