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) + \

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

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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,$\sqrt{1 + \sin(2x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|$. 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 $x$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

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

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