If I_n = int_0^{pi / 2} sin^n(x) dx and I_n = | vedclass

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(B) We are given $I_n = \int_0^{\pi / 2} \sin^n(x) dx$. Using integration by parts,let $u = \sin^{n-1}(x)$ and $dv = \sin(x) dx$. Then $du = (n-1) \si

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$\frac{n-1}{n}$

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(B) We are given $I_n = \int_0^{\pi / 2} \sin^n(x) dx$. Using integration by parts,let $u = \sin^{n-1}(x)$ and $dv = \sin(x) dx$. Then $du = (n-1) \sin^{n-2}(x) \cos(x) dx$ and $v = -\cos(x)$. $I_n = [-\sin^{n-1}(x) \cos(x)]_0^{\pi / 2} + \int_0^{\pi / 2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$. Since $\cos(\pi/2) = 0$ and $\sin(0) = 0$,the boundary term is $0$. $I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) (1 - \sin^2(x)) dx$. $I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) dx - (n-1) \int_0^{\pi / 2} \sin^n(x) dx$. $I_n = (n-1) I_{n-2} - (n-1) I_n$. $I_n + (n-1) I_n = (n-1) I_{n-2}$. $n I_n = (n-1) I_{n-2}$. $I_n = \frac{n-1}{n} I_{n-2}$. Comparing this with $I_n = k I_{n-2}$,we get $k = \frac{n-1}{n}$. Hence,option $B$ is correct.

If $I_n = \int_0^{\pi / 2} \sin^n(x) dx$ and $I_n = (k) I_{n-2}$,then what will be the value of $k$?

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