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Let $f(x)$ be an indefinite integral of $\sin ^{2} x .$ Consider the following statements:
Statement $1:$ The function $f(x)$ satisfies $f(x+\pi)$ $=f(\mathrm{x})$ for all real $\mathrm{x}$. Statement $2: \sin ^{2}(x+\pi)=\sin ^{2} x$ for all real $x$. Which one of the following is correct in respect of the above statements?
Options:
Statement $1:$ The function $f(x)$ satisfies $f(x+\pi)$ $=f(\mathrm{x})$ for all real $\mathrm{x}$. Statement $2: \sin ^{2}(x+\pi)=\sin ^{2} x$ for all real $x$. Which one of the following is correct in respect of the above statements?
Solution:
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The correct answer is:
Both the statements are true but $\mathrm{Statement} 2$ is not the correct explanation of Statement 1
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int \sin ^{2} \mathrm{x} \cdot \mathrm{dx}=\int \frac{1-\cos 2 \mathrm{x}}{2} \mathrm{~d} \mathrm{x} \\ &=\frac{1}{2} \mathrm{x}-\frac{\sin 2 \mathrm{x}}{2} \cdot \frac{1}{2}+\mathrm{c}=\frac{1}{2} \mathrm{x}-\frac{1}{4} \sin 2 \mathrm{x}+\mathrm{c} \\ \text { (i) } & \mathrm{f}(\pi+\mathrm{x})=\frac{1}{2}(\pi+\mathrm{x})-\frac{1}{4} \sin 2(\pi+\mathrm{x}) \\ &=\frac{1}{2} \pi+\frac{1}{2} \mathrm{x}-\frac{1}{2} \sin (2 \pi+2 \mathrm{x})+\mathrm{c} \\ &=\frac{\mathrm{x}}{2}-\frac{1}{2} \sin 2 \mathrm{x}+\mathrm{c}=\mathrm{f}(\mathrm{x}) \end{aligned}$
So, statement 1 is true.
(ii) $\sin ^{2}(\pi+x)=\sin ^{2} x$
$(-\sin x)^{2}=\sin ^{2} x$
$\Rightarrow \sin ^{2} x=\sin ^{2} x$
So, statement 2 is true
So, statement 1 is true.
(ii) $\sin ^{2}(\pi+x)=\sin ^{2} x$
$(-\sin x)^{2}=\sin ^{2} x$
$\Rightarrow \sin ^{2} x=\sin ^{2} x$
So, statement 2 is true
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