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

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(A) We have $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$. Using integration by parts,$I_n = [x^n \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} n x^{n-1} \sin x

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

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(A) We have $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$. Using integration by parts,$I_n = [x^n \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} n x^{n-1} \sin x \, dx$. $I_n = (\frac{\pi}{2})^n - n \int_0^{\pi / 2} x^{n-1} \sin x \, dx$. Applying integration by parts again to the integral,$\int_0^{\pi / 2} x^{n-1} \sin x \, dx = [x^{n-1} (-\cos x)]_0^{\pi / 2} - \int_0^{\pi / 2} (n-1) x^{n-2} (-\cos x) \, dx = 0 + (n-1) I_{n-2}$. Substituting this back,$I_n = (\frac{\pi}{2})^n - n(n-1) I_{n-2}$. Thus,$I_n + n(n-1) I_{n-2} = (\frac{\pi}{2})^n$. Dividing by $n!$,we get $\frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} = \frac{(\pi/2)^n}{n!}$. Summing from $n=2$ to $\infty$,$\sum_{n=2}^{\infty} \frac{(\pi/2)^n}{n!} = (\sum_{n=0}^{\infty} \frac{(\pi/2)^n}{n!}) - \frac{(\pi/2)^0}{0!} - \frac{(\pi/2)^1}{1!} = e^{\pi / 2} - 1 - \frac{\pi}{2}$.

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

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