Trigonometric and Exponential Integration 
If $\displaystyle I_{n} =\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx.$  Then $\displaystyle \frac{I_{2018}}{I_{2016}}$ is 

Try: using by parts
$$I_{n}=\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx$$
$$I_{n}=-\cos^{n}(x)\cdot e^{-x}\bigg|^{\infty}_{\frac{\pi}{2}}-n\int^{\infty}_{\frac{\pi}{2}}\cos^{n-1}(x)\sin(x)\cdot e^{-x}dx$$
$$I_{n}=n\int^{\infty}_{\frac{\pi}{2}}\cos^{n-1}(x)\sin(x)\cdot e^{-x}dx$$
Could some help me to solve it, Thanks

As you say,
$$
\frac{I_n}{n} = - \int_{\pi/2}^{\infty} \mathrm{e}^{-x} \cos^{n-1}(x)\sin(x)\ \mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
\begin{align*}
\frac{\mathrm{d} }{\mathrm{d} x} \, \cos^{n-1}(x)\sin(x) &= \cos^n(x) - (n-1)\cos^{n-2}(x)\color{blue}{\sin^2(x)} \\
& = \cos^n(x) - (n-1)\cos^{n-2}(x)\left[\color{blue}{1-\cos^2(x)}\right] \\ 
& = \color{red}{\cos^n(x)} - (n-1) \cos^{n-2}(x) + (n\color{red}{-1})\cos^n(x) \\
& = n \cos^n(x) - (n-1) \cos^{n-2}(x).
\end{align*}
Then the whole integral becomes
$$
\frac{I_n}{n} = - \displaystyle \color{red}{\left[ -\mathrm{e}^{-x} \cos^{n-1}(x)\sin(x)\right]^{\infty}_{\pi/2}} -  \int_{\pi/2}^{\infty} \mathrm{e}^{-x} \left[ n \cos^n(x) - (n-1) \cos^{n-2}(x)\right]\ \mathrm{d}x
$$
with the red term being zero, we obtain
\begin{align*}
\frac{I_n}{n} &= -nI_n + (n-1)I_{n-2}, \\
I_n & = - n^2 I_n + n(n-1)I_{n-2}, \\
(n^2 + 1)I_n &= n(n-1)I_{n-2}, \\
\frac{I_n}{I_{n-2}} &= \frac{n(n-1)}{n^2+1}, \\
\end{align*}
In particular, 
$$
\frac{I_{2018}}{I_{2016}} = \frac{2018 \times 2017}{2018^2+1} = \frac{4070306}{4072325}.
$$
