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, $$ \frac{I_{2018}}{I_{2016}} = \frac{2018 \times 2017}{2018^2+1} = \frac{4070306}{4072325}. $$
For brevity we set
$$f_{c,s}(x):= e^{-x}\cos^c(x)\sin^s(x),$$ $$F_{c,s}(x):=\int f_{c,s}(x)\,dx.$$
Then by parts, integrating on $e^{-x}$, $$F_{c,0}=-f_{c,0}-cF_{c-1,1}$$ and $$F_{c-1,1}=-f_{c-1,1}-(c-1)F_{c-2,2}+F_{c,0}.$$
Using $\sin^2x=1-\cos^2x$ (i.e. $F_{c-2,2}=F_{c-2,0}-F_{c,0}$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_{c,0}=c(c-1)F_{c-2,0}.$$