Let f(x) = frac{sin x}{x},then int_{0}^{frac{ | vedclass

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(D) Given $f(x) = \frac{\sin x}{x}$. We need to evaluate $I = \int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx$. Substituting the func

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$\frac{1}{\pi} \int_{0}^{\pi} f(x) dx$

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(D) Given $f(x) = \frac{\sin x}{x}$. We need to evaluate $I = \int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx$. Substituting the function,we get $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{x} \cdot \frac{\sin(\frac{\pi}{2} - x)}{\frac{\pi}{2} - x} dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{x(\frac{\pi}{2} - x)} dx$. This simplifies to $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2x}{2x(\frac{\pi}{2} - x)} dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2x}{x(\pi - 2x)} dx$. Let $t = 2x$,then $dt = 2dx$ or $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$. $I = \int_{0}^{\pi} \frac{\sin t}{t(\pi - t)} \cdot \frac{dt}{2} = \frac{1}{2} \int_{0}^{\pi} \frac{\sin t}{t(\pi - t)} dt$. Using partial fractions,$\frac{1}{t(\pi - t)} = \frac{1}{\pi} \left( \frac{1}{t} + \frac{1}{\pi - t} \right)$. $I = \frac{1}{2\pi} \int_{0}^{\pi} \sin t \left( \frac{1}{t} + \frac{1}{\pi - t} \right) dt = \frac{1}{2\pi} \left( \int_{0}^{\pi} \frac{\sin t}{t} dt + \int_{0}^{\pi} \frac{\sin t}{\pi - t} dt \right)$. In the second integral,let $u = \pi - t$,then $du = -dt$. The integral becomes $\int_{0}^{\pi} \frac{\sin u}{u} du$. Thus,$I = \frac{1}{2\pi} \left( 2 \int_{0}^{\pi} \frac{\sin t}{t} dt \right) = \frac{1}{\pi} \int_{0}^{\pi} f(t) dt$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

Let $f(x) = \frac{\sin x}{x}$,then $\int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx =$

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