Question

If $I_n=\underset{\pi /2}{\overset{\infty }{\int }}{e}^{-x}{\cos }^{n}xdx]$, then $\frac{I_{2018}}{I_{2016}}=$

• A. $\frac{2018\times 2019}{(2017{)}^2+1}$
• B. $\frac{2018\times 2017}{(2018{)}^2+1}$
• C. $\frac{(2018)(2016)}{(2017{)}^2+1}$
• D. $\frac{(2018)(2017)}{(2019{)}^2+1}$

Answer 1

Answer: (B)

We have,
$I_n=\underset{\pi /2}{\overset{\infty }{\int }}{e}^{-x}{\cos }^{n}xdx$
Integration by parts
$I_n={\left[-e^{-x}{\cos }^{n}x\right]}_{\pi /2}^{\infty }$
$-{\int }_{\pi /2}^{\infty }-e^{-x}\cdot n{\cos }^{n-1}x(-\sin x)dx$
$=0-\underset{\pi /2}{\overset{\infty }{\int }}n{e}^{-x}{\cos }^{n-1}x\sin xdx$
By integration parts again, we get
$=-n[{\left(-e^{-x}{\cos }^{n-1}x\sin x\right]}_{\pi /2}^{\infty }$
$-\underset{\pi /2}{\overset{\infty }{\int }}-e^{-x}\left({\cos }^{n}x-(n-1){\cos }^{n-2}x{\sin }^{2}x\right)dx]$
$=-n[0+\underset{\pi /2}{\overset{\infty }{\int }}(e^{-x}{\cos }^{n}x-(n-1)e^{-x}$
${\cos }^{n-2}x\left(1-{\cos }^{2}x\right)dx]$
$\left[\because {\sin }^{2}x=1-{\cos }^{2}x\right]$
$=-n[\underset{\pi /2}{\overset{\infty }{\int }}{e}^{-x}{\cos }^{n}xdx-(n-1)$
$\underset{\pi /2}{\overset{\infty }{\int }}{e}^{-x}{\cos }^{n-2}xdx+(n-1){\int }_{\pi /2}^{\infty }{e}^{-x}{\cos }^{n}xdx]$
$=-n\left[I_n-(n-1)I_{n-2}+(n-1)I_n\right]$
$\Rightarrow I_n=-n\left[n{I}_n-(n-1)I_{n-2}\right]$
$\Rightarrow I_n=-n^2{I}_n+n(n-1)I_{n-2}$
$\Rightarrow I_n+n^2{I}_n=n(n-1)I_{n-2}$
$\Rightarrow I_n\left(n^2+1\right)=n(n-1)I_{n-2}$
$\Rightarrow \frac{I_n}{I_{n-2}}=\frac{n(n-1)}{n^2+1}$
Put $n=2018$
$\frac{I_{2018}}{I_{2016}}=\frac{2018(2018-1)}{(2018{)}^2+1}=\frac{2018\times 2017}{(2018{)}^2+1}$