Question

Let $I_n={\int }_0^{\pi /2}{x}^n\cos xdx,$ where n is a non negative integer. Then $\sum _{n=2}^{\infty }\left(\frac{I_n}{n!}+\frac{I_{n-2}}{(n-2)!}\right)$ equal.

• A. $e^{\pi /2}-1-\frac{\pi }{2}$
• B. $e^{\pi /2}-1$
• C. $e^{\pi /2}-\frac{\pi }{2}$
• D. $e^{\pi /2}$

Answer 1

The correct option is A $e^{\pi /2}-1-\frac{\pi }{2}$

$I_n=\underset{0}{\overset{\pi /2}{\int }}{x}^n\cos xdx=\left(x^n\sin x\right){|}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}n{x}^{n-1}\sin xdx\left(x^n\sin x\right){|}_0^{\pi /2}=\left(\frac{\pi }{2}{)}^n\underset{0}{\overset{\pi /2}{\int }}{x}^{n-1}\sin xdx=\left(x^{n-1}\right(-\cos x\left)\right){|}_0^{\pi /2}+\underset{0}{\overset{\pi /2}{\int }}(n-1)x^{n-2}\cos xdx\right(x^{n-1}(-\cos x)){|}_0^{\pi /2}=0\underset{0}{\overset{\pi /2}{\int }}{x}^{n-2}\cos xdx=I_{n-2}$

Therefore, $I_n=(\frac{\pi }{2}{)}^n-n(n-1)I_{n-2}\Rightarrow \frac{I_n}{n!}+\frac{I_{n-2}}{(n-2)!}=\frac{(\frac{\pi }{2}{)}^n}{n!}$

Thus, $\sum _{n=2}^{\infty }(\frac{I_n}{n!}+\frac{I_{n-2}}{(n-2)!})=\sum _{n=0}^{\infty }\frac{(\frac{\pi }{2}{)}^n}{n!}-1-\frac{\pi }{2}=e^{\pi /2}-\frac{\pi }{2}-1$