Question

Let $I_n={\int }_0^{\infty }{e}^{-x}(sinx{)}^{n}dx,nϵN,n>1$ then $\frac{I_{2008}}{I_{2006}}$ equals

• A. $\frac{2007\times 2006}{{2008}^2+1}$
• B. $\frac{2008\times 2007}{{2008}^2+1}$
• C. $\frac{2006\times 2004}{{2008}^2-1}$
• D. $\frac{2008\times 2007}{{2008}^2-1}$

Answer 1

The correct option is B $\frac{2008\times 2007}{{2008}^2+1}$

$In={\int }_0^{\infty }{e}^{-x}(sinx{)}^{n}dx=[si{n}^{n}x(-e^{-x}){]}_0^{\infty }+{\int }_0^{\infty }nsi{n}^{n-1}xcosx{e}^{-x}dx=0+n{\int }_0^{\infty }\left(si{n}^{n-1}xcosx\right)e^{-x}dx=n\left[\right(si{n}^{n-1}xcosx\left)\right(-e^{-x}){]}_0^{\infty }-n{\int }_0^{\infty }\{-si{n}^{n}x+(n-1)si{n}^{n-2}xco{s}^{2}x\left\}\right(-e^{-x})dx=0+n{\int }_0^{\infty }{e}^{-x}\{-si{n}^{n}x+(n-1)si{n}^{n-2}x(1-si{n}^{2}x)\}dx=n{\int }_0^{\infty }{e}^{-x}\{(n-1)si{n}^{n-2}x-nsi{n}^{n}x\}dx=n(n-1)I_{n-2}-n^2{I}_n(1+n^2)I_n-2-n^2{I}_n$
we have $(1+n^2)I_n=\frac{n(n-1)}{I_{n-2}}$
then $\frac{I_n}{I_{n-2}}=\frac{n(n-1)}{n^2+1}$