Let F(x) be an indefinite integral of sin ^2 | vedclass

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(D) Given $F(x) = \int \sin^2 x \, dx$.Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have:$F(x) = \int \frac{1 - \cos 2x}{2} \, dx = \frac{

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Statement -$1$ is False,Statement -$2$ is True

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(D) Given $F(x) = \int \sin^2 x \, dx$.Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have:$F(x) = \int \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2} x - \frac{1}{4} \sin 2x + C$.Now,check $F(x + \pi)$:$F(x + \pi) = \frac{1}{2}(x + \pi) - \frac{1}{4} \sin(2(x + \pi)) + C$$F(x + \pi) = \frac{1}{2}x + \frac{\pi}{2} - \frac{1}{4} \sin(2x + 2\pi) + C$$F(x + \pi) = \frac{1}{2}x + \frac{\pi}{2} - \frac{1}{4} \sin 2x + C = F(x) + \frac{\pi}{2}$.Since $F(x + \pi) 
eq F(x)$,Statement-$1$ is False.For Statement-$2$,we know $\sin(x + \pi) = -\sin x$,so $\sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x$. Thus,Statement-$2$ is True.Therefore,Statement-$1$ is False and Statement-$2$ is True.

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

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because
$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

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Let $F(x)$ be an indefinite integral of $\sin ^2 x$.$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.