Question

If $\left.I_{n}=\int\limits_{\pi / 2}^{\infty} e^{-x} \cos ^{n} \,x \,d x\right]$, 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,
$I_{n}=\int\limits_{\pi / 2}^{\infty} e^{-x} \cos ^{n} x d x$
Integration by parts
$I_{n}=\left[-e^{-x} \cos ^{n} x\right]_{\pi / 2}^{\infty} $
$-\int_{\pi / 2}^{\infty}-e^{-x} \cdot n \cos ^{n-1} x(-\sin x) d x$
$=0-\int\limits_{\pi / 2}^{\infty} n e^{-x} \,\cos ^{n-1} \,x \,\sin x \,d x$
By integration parts again, we get
$=-n\left[\left(-e^{-x} \cos ^{n-1} x \sin x\right]_{\pi / 2}^{\infty}\right. $
$\left.-\int\limits_{\pi / 2}^{\infty}-e^{-x}\left(\cos ^{n} x-(n-1) \cos ^{n-2} x \sin ^{2} x\right) d x\right]$
$=-n\left[0+\int\limits_{\pi / 2}^{\infty}\left(e^{-x} \cos ^{n} x-(n-1) e^{-x}\right.\right.$
$\left.\cos ^{n-2} x\left(1-\cos ^{2} x\right) d x\right]$
$\left[\because \sin ^{2} x=1-\cos ^{2} \,x\right]$
$=-n\left[\int\limits_{\pi / 2}^{\infty} e^{-x} \cos ^{n} x d x-(n-1)\right.$
$\left.\int\limits_{\pi / 2}^{\infty} e^{-x} \cos ^{n-2} x d x+(n-1) \int_{\pi / 2}^{\infty} e^{-x} \cos ^{n} \,x d x\right]$
$ =-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} $