Let I_n=int_0^{pi / 2} x^n cos x d x, where n | vedclass

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Question

(a)

We have,

$I_n=\int \limits_0^{\pi / 2} x^n \cos x d x$

$\Rightarrow \quad I_n =\left[x^n \sin x\right]_0^{\pi / 2}-\int \limits_0^{\pi /

Vedclass · Accepted Answer

$e^{\pi / 2}-1-\frac{\pi}{2}$

Vedclass · Answer

(a)

We have,

$I_n=\int \limits_0^{\pi / 2} x^n \cos x d x$

$\Rightarrow \quad I_n =\left[x^n \sin xight]_0^{\pi / 2}-\int \limits_0^{\pi / 2} n x^{n-1} \sin x d x$

$\Rightarrow \quad I_n =\left(\frac{\pi}{2}ight)^n-\left[n x^{n-1}(-\cos x)ight]_0^{\pi / 2}$

$\Rightarrow \quad I_n =\left(\frac{\pi}{2}ight)^n-n(n-1) I_{n-2}^{\pi / 2} n(n-1) x^{n-2} \cos x d x$

$\Rightarrow \quad I_n+n n(n-1) I_{n-2}=\left(\frac{\pi}{2}ight)^n$

$	ext { Now, } \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !}+\frac{I_{n-2}}{(n-2) !}ight]$

$\Rightarrow \quad \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !} n(n-1) I_{n-2}ight]$

$\sum \limits_{n=2}^{\infty} \frac{\left(\frac{\pi}{2}ight)^n}{n !}=e^{\pi / 2}-1-\frac{\pi}{2}$

Let $I_n=\int_0^{\pi / 2} x^n \cos x d x$, where $n$ is a non-negative integer. Then, $\sum \limits_{n=2}^{\infty}\left(\frac{I_n}{n !}+\frac{I_n-2}{(n-2) !}\right)$ equals

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