Asymptotics of $\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du$ as $r \to \infty$. Can someone explain the following asymptotics, through Watson’s Lemma or through another argument?

\begin{align}
\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{\pi}{2(r+1)^2},\qquad r\to \infty.
\end{align}

 A: By just integrating by parts twice one gets, as $r \to \infty$,
$$
\begin{align}
&\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du
\\\\&=\left[ \frac{e^{-(r+1)u}}{-(r+1)}\cos\left(\frac{\pi}{2}e^{-u}\right)\right]_0^\infty
+\frac{\pi}{2(r+1)}\int^\infty_0 e^{-(r+2)u}\sin\left(\frac{\pi}{2}e^{-u}\right)\,du
\\\\&=\color{red}{0}+\frac{\pi}{2(r+1)}\left(\left[ \frac{e^{-(r+2)u}}{-(r+2)}\sin\left(\frac{\pi}{2}e^{-u}\right)\right]_0^\infty
-\frac{\pi}{2(r+2)}\int^\infty_0 e^{-(r+3)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \right)
\\\\&=\frac{\pi}{2(r+1)(r+2)}+o\left(\frac1{(r+1)^2}\right)
\\\\&=\frac{\pi}{2(r+1)^2}+o\left(\frac1{(r+1)^2}\right).
\end{align}
$$
A: If we want to make it match the formula for Watson's lemma on wikipedia, we start by writing
$$
\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du = \int^\infty_0 e^{-ru} e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du = \int^\infty_0 e^{-ru} \varphi(u)\,du,
$$
where
$$
\varphi(u) := e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right).
$$
Now $\varphi(0) = 0$ and $\varphi'(0) = \pi/2$, so we can write
$$
\varphi(u) = ug(u),
$$
where
$$
g(u) = u^{-1} e^{-u}\cos\left(\frac{\pi}{2}e^{-u}\right)
$$
and $g(0) = \pi/2 \neq 0$. In the notation of wikipedia we have $\lambda = 1$.
Thus, taking only the first term ($n=0$) of the asymptotic expansion from the article, we get
$$
\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{g^{(0)}(0)\, \Gamma(1+0+1)}{0!\, r^{1+0+1}} = \frac{\pi}{2r^2}
$$
as $r \to \infty$ since
$$
g^{(0)}(0) = g(0) = \pi/2 \qquad \text{and} \qquad \Gamma(2) = 1! = 1.
$$

To conclude that
$$
\int^\infty_0 e^{-(r+1)u}\cos\left(\frac{\pi}{2}e^{-u}\right)\,du \sim \frac{\pi}{2(r+1)^2}
$$
we just need to note that
$$
\frac{\pi}{2r^2} \sim \frac{\pi}{2(r+1)^2}
$$
as $r \to \infty$.
