Question

For $n\in N$, define $I_n={\int }_0^{\infty }{e}^{-x}(sinx{)}^{m}dx$. Then $\frac{I_{2008}}{I_{2006}}$ equals:

• A. $\frac{2007.2006}{{2007}^2+1}$
  
• B. $\frac{2007.2008}{{2008}^2+1}$
  
• C. $\frac{2007}{2008}$
  
• D. $\frac{2007}{2006}$

Answer 1

The correct option is B $\frac{2007.2008}{{2008}^2+1}$


$I_n=\frac{n(n-1)I_n-2}{n^2+1},\frac{I_n}{I_n-2}=\frac{n(n-1)}{n^2+1},n=2008$