The basic idea is that the maximum contribution of the integral comes from a neighborhood of $t=0$, and near there we have $t^2+2t^4 \approx t^2$. This problem is particularly nice because we can do everything explicitly. I'll do the calculation in two steps (two changes of variables) to illustrate what's going on.
Start with the change of variables $t^2+2t^4 = s^2$, where $s \geq 0$. This gives
$$
t=\frac{1}{2}\sqrt{-1+\sqrt{1+8s^2}},
$$
so that the integral becomes
$$
\int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}\,dt = \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds,
$$
where
$$
f(s) = \frac{2^{1/4}s^{1/4}}{\sqrt{1+8s^2}\left(-1+\sqrt{1+8s^2}\right)^{1/8}} = 1 - \frac{15}{4}s^2 + \frac{713}{32}s^4 + \cdots.
$$
Now we can make the second change of variables $s^2 = r$ to put the integral into a form where we can directly apply Watson's lemma. Indeed, this gives
$$
\int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds = \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr,
$$
where
$$
g(r) = \frac{1}{2}f\left(\sqrt{r}\right) = \frac{1}{2} - \frac{15}{8}r + \frac{713}{64}r^2 + \cdots.
$$
Finally
$$
\begin{align*}
\int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr &\approx \sum_{n=0}^{\infty} \frac{g^{(n)}(0) \Gamma(n+7/8)}{n! x^{n+7/8}} \\
&= \frac{1}{2}\Gamma\left(\frac{7}{8}\right) x^{-7/8} - \frac{15}{8}\Gamma\left(\frac{15}{8}\right)x^{-15/8} + O\left(x^{-23/8}\right)
\end{align*}
$$
as $x \to \infty$, by Watson's lemma.
You can make the change of variables $t = \operatorname{arcsinh} \sqrt u$ to reduce the integral to the standard case. Essentially, this holds:
$$\int_0^1 e^{-x \sinh^4 t} \sin t\,dt \sim
\int_0^\infty e^{-x t^4} t \,dt =
\frac 1 2 \int_0^\infty e^{-x (t^2)^2} d(t^2).$$
Best Answer
Plot integrand for few values of $x$:
It is apparent that the maximum shifts closer to the origin as $x$ grows.
Let's rewrite the integrand as follows: $$ \int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t = \int_0^{\pi/2} \exp\left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) \mathrm{d}t $$ The maximum of the integrand is determined by $$ 0 = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) = \cot(t) \left( \frac{1}{2} - 4 x \sin^4(t)\right) $$ that is at $t_\ast = \arcsin\left((8 x)^{-1/4}\right)$. Then using Laplace's method: $$ \int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx \int_{0}^{\pi/2} \exp\left(\phi(t_\ast) + \frac{1}{2} \phi^{\prime\prime}(t_\ast) (t-t_\ast)^2 \right) \mathrm{d}t = \exp\left(\phi(t_\ast)\right) \sqrt{\frac{2\pi}{-\phi^{\prime\prime}(t_\ast)}} $$ Easy algebra gives $\exp\left(\phi(t_\ast)\right) = (8 \mathrm{e} x)^{-1/8}$, $-\phi^{\prime\prime}(t_\ast) = 4 \sqrt{2 x} - 2$, giving $$ \int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx (8 \mathrm{e} x)^{-1/8} \sqrt{ \frac{\pi}{2 \sqrt{2 x} -1}} $$