If a_n = int_0^{frac{pi}{2}} frac{sin^2 nx}{s | vedclass

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(C) Consider the difference $a_{n+1} - a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2((n+1)x) - \sin^2(nx)}{\sin x} dx$. Using the identity $\sin^2 A - \si

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Harmonic progression

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(C) Consider the difference $a_{n+1} - a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2((n+1)x) - \sin^2(nx)}{\sin x} dx$. Using the identity $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B)$,we get $\sin^2((n+1)x) - \sin^2(nx) = \sin(x)\sin((2n+1)x)$. Thus,$a_{n+1} - a_n = \int_0^{\frac{\pi}{2}} \frac{\sin(x)\sin((2n+1)x)}{\sin x} dx = \int_0^{\frac{\pi}{2}} \sin((2n+1)x) dx$. Evaluating the integral: $a_{n+1} - a_n = \left[ -\frac{\cos((2n+1)x)}{2n+1} \right]_0^{\frac{\pi}{2}} = -\frac{1}{2n+1} (\cos((2n+1)\frac{\pi}{2}) - \cos(0)) = -\frac{1}{2n+1} (0 - 1) = \frac{1}{2n+1}$. For $n=1$,$a_2 - a_1 = \frac{1}{2(1)+1} = \frac{1}{3}$. For $n=2$,$a_3 - a_2 = \frac{1}{2(2)+1} = \frac{1}{5}$. For $n=3$,$a_4 - a_3 = \frac{1}{2(3)+1} = \frac{1}{7}$. The sequence is $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots$. Since the reciprocals $3, 5, 7, \dots$ are in Arithmetic Progression,the sequence is in Harmonic Progression.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

If $a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} dx$,then $a_2 - a_1, a_3 - a_2, a_4 - a_3$ are in

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