Which of the following are true?

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(D) $1$. For option $A$: Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $I = \int_{a}^{\pi-a} x f(\sin x) dx$. Replacing

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All of the above

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(D) $1$. For option $A$: Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $I = \int_{a}^{\pi-a} x f(\sin x) dx$. Replacing $x$ with $a + (\pi - a) - x = \pi - x$,we get $I = \int_{a}^{\pi-a} (\pi - x) f(\sin(\pi - x)) dx = \int_{a}^{\pi-a} (\pi - x) f(\sin x) dx$. Adding the two expressions: $2I = \int_{a}^{\pi-a} \pi f(\sin x) dx$,so $I = \frac{\pi}{2} \int_{a}^{\pi-a} f(\sin x) dx$. This is true.$2$. For option $B$: If $f(x)^2$ is an even function,then $\int_{-a}^{a} f(x)^2 dx = 2 \int_{0}^{a} f(x)^2 dx$. Since $f(x)^2$ is always non-negative and symmetric for any function $f(x)$,the property holds. This is true.$3$. For option $C$: Using the property $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$ if $f(x+T) = f(x)$,here $f(\cos^2 x)$ has a period of $\pi$. Thus,$\int_{0}^{n\pi} f(\cos^2 x) dx = n \int_{0}^{\pi} f(\cos^2 x) dx$. This is true.$4$. Since all statements are true,the correct option is $D$.

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