If I_n = int_{pi / 2}^{infty} e^{-x} cos^n x | vedclass

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(B) We have $I_n = \int_{\pi / 2}^{\infty} e^{-x} \cos^n x \, dx$.Using integration by parts, let $u = \cos^n x$ and $dv = e^{-x} dx$. Then $du = n \c

Vedclass · Accepted Answer

$\frac{2018 	imes 2017}{(2018)^2+1}$

Vedclass · Answer

(B) We have $I_n = \int_{\pi / 2}^{\infty} e^{-x} \cos^n x \, dx$.Using integration by parts, let $u = \cos^n x$ and $dv = e^{-x} dx$. Then $du = n \cos^{n-1} x (-\sin x) dx$ and $v = -e^{-x}$.$I_n = [-e^{-x} \cos^n x]_{\pi / 2}^{\infty} - \int_{\pi / 2}^{\infty} (-e^{-x}) (n \cos^{n-1} x) (-\sin x) dx$$I_n = 0 - n \int_{\pi / 2}^{\infty} e^{-x} \cos^{n-1} x \sin x \, dx$.Applying integration by parts again to the integral $\int_{\pi / 2}^{\infty} e^{-x} \cos^{n-1} x \sin x \, dx$ with $u = \cos^{n-1} x \sin x$ and $dv = e^{-x} dx$:$I_n = -n [(-e^{-x} \cos^{n-1} x \sin x)_{\pi / 2}^{\infty} - \int_{\pi / 2}^{\infty} (-e^{-x}) ((n-1) \cos^{n-2} x (-\sin^2 x) + \cos^n x) dx]$$I_n = -n [0 + \int_{\pi / 2}^{\infty} e^{-x} (-(n-1) \cos^{n-2} x (1 - \cos^2 x) + \cos^n x) dx]$$I_n = -n [-(n-1) I_{n-2} + (n-1) I_n + I_n]$$I_n = -n [n I_n - (n-1) I_{n-2}]$$I_n = -n^2 I_n + n(n-1) I_{n-2}$$I_n(1 + n^2) = n(n-1) I_{n-2}$$\frac{I_n}{I_{n-2}} = \frac{n(n-1)}{n^2+1}$.For $n = 2018$, we get $\frac{I_{2018}}{I_{2016}} = \frac{2018 	imes 2017}{(2018)^2+1}$.

If $I_n = \int_{\pi / 2}^{\infty} e^{-x} \cos^n x \, dx$, then $\frac{I_{2018}}{I_{2016}} = $

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