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.

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 $x$.