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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $x$.

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

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

Similar Questions

$\int \frac{x+\sin x}{1+\cos x} d x=$

$\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=$

If $\int \frac{\cos (13 x)-\cos (14 x)}{1+2 \cos (9 x)} d x=\frac{\sin (4 x)}{a}-\frac{\sin (5 x)}{b}+c$,then $a^b=$

$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module