Periodicity of the Indefinite Integral of Sine Squared

To determine the validity of the given statements, let us analyze each of them:

Evaluation of Statement-2:
We know the trigonometric identity:
$\sin(x + \pi) = -\sin x$
Squaring both sides gives:
$\sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x \quad \text{for all real } x$
Thus, $\sin^2 x$ is a periodic function with period $\pi$.
Therefore, Statement-2 is True.

Evaluation of Statement-1:
Given that $F(x)$ is an indefinite integral of $\sin^2 x$, we have:
$F(x) = \int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$
where $C$ is an arbitrary constant of integration.

Now, let us calculate $F(x + \pi)$:
$F(x + \pi) = \frac{x + \pi}{2} - \frac{\sin(2(x + \pi))}{4} + C$
$F(x + \pi) = \frac{x}{2} + \frac{\pi}{2} - \frac{\sin(2x + 2\pi)}{4} + C$
Since $\sin(2x + 2\pi) = \sin 2x$, this simplifies to:
$F(x + \pi) = \left(\frac{x}{2} - \frac{\sin 2x}{4} + C\right) + \frac{\pi}{2} = F(x) + \frac{\pi}{2}$
Alternatively, by the Fundamental Theorem of Calculus:
$F(x + \pi) - F(x) = \int_{x}^{x + \pi} \sin^2 t \, dt = \frac{\pi}{2} \neq 0$
Since $F(x + \pi) = F(x) + \frac{\pi}{2} \neq F(x)$ for all real $x$, $F(x)$ is not periodic with period $\pi$.
Therefore, Statement-1 is False.

Conclusion:
Statement-1 is False, and Statement-2 is True.

Hence, the correct option is (D).