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

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(C) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Alternatively,$1 + \sin x = 1 + \cos(\pi/2

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$\sqrt{2} \log 	an \left( \frac{\pi}{8} + \frac{x}{4} ight) + c$

Vedclass · Answer

(C) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Alternatively,$1 + \sin x = 1 + \cos(\pi/2 - x) = 2\cos^2(\pi/4 - x/2)$.Thus,$\sqrt{1 + \sin x} = \sqrt{2} \cos(\pi/4 - x/2) = \sqrt{2} \sin(\pi/4 + x/2)$.The integral becomes $\int \frac{1}{\sqrt{2} \sin(\pi/4 + x/2)} \, dx$.This is equal to $\frac{1}{\sqrt{2}} \int \csc(\pi/4 + x/2) \, dx$.Using the formula $\int \csc 	heta \, d	heta = \log 	an(	heta/2) + c$,we get:$\frac{1}{\sqrt{2}} \cdot 2 \log 	an \left( \frac{\pi/4 + x/2}{2} ight) + c = \sqrt{2} \log 	an \left( \frac{\pi}{8} + \frac{x}{4} ight) + c$.

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

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