int(sqrt{1+sin (2 x)}) d x= | vedclass

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(D) We know that $1 + \sin(2x) = \sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = (\sin(x) + \cos(x))^2$. Therefore,$\sqrt{1 + \sin(2x)} = \sqrt{(\sin(x) + \

Vedclass · Accepted Answer

$	ext{Can be option } B 	ext{ or } C 	ext{ depending on the value of } x$

Vedclass · Answer

(D) We know that $1 + \sin(2x) = \sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = (\sin(x) + \cos(x))^2$. Therefore,$\sqrt{1 + \sin(2x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|$. Thus,the integral becomes $I = \int |\sin(x) + \cos(x)| dx$. This integral depends on the sign of $(\sin(x) + \cos(x))$: If $\sin(x) + \cos(x) \geq 0$,then $I = \int (\sin(x) + \cos(x)) dx = -\cos(x) + \sin(x) + C = \sin(x) - \cos(x) + C$. If $\sin(x) + \cos(x) Since the result depends on the interval of $x$,the correct answer is that it can be option $B$ or $C$ depending on the value of $x$.

$\int(\sqrt{1+\sin (2 x)}) d x=$

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