$$ \int_0^{1000} \exp(1000\sin(x)/x)\,{\rm d}x$$
Solve this integral of a sharply peaked function without a calculator.
I was told to not do an expansion of the sharp function, but of its gently varying logarithm. Once that was done I should then see a function that I can expand. Once expanded put it back into the integral for the result.
$$ \ln [\exp(1000\sin(x)/x)] \ $$ $$ =[1000\sin(x)/x)] \ $$ Now the expansion of $\sin(x)/x$:
$$ 1 -x^2/3! + x^4/5!+\ldots +(-1)^nx^{2n+1} / (2n+1)! $$
$$ 1000 \int_0^{1000} \sum (-1)^nx^{2n+1} / (2n+1)!\,{\rm d}x$$
$$ 1000 \int_0^{1000} \sum (-1)^n x^{2n+2} / (2n+1)(2n+2)!\, {\rm d}x$$
I don't understand why when I put the integral into wolfram I get an answer of $1.35\cdot 10^{433}$, however when I put the integral into my calculator I get an answer of $1.14\cdot 10^9$.
What you are asked to do is to derive what is called Laplace's method, which is a way of estimating integrals on the form $\int e^{Mf(x)}{\rm d}x$ where $M$ is some large number, for the special case of $f(x) = \frac{\sin(x)}{x}$.
As you have done we expand the $\log$ of the exponential in a Taylor series to second order $$1000\frac{\sin(x)}{x} = 1000 - \frac{1000}{6}x^{2} + \mathcal{O}(x^3)$$ so your integrand close to $x=0$ is given by $$e^{1000\frac{\sin(x)}{x}} = e^{1000 - \frac{1000}{6}x^{2} + \mathcal{O}(x^3)}$$ Note that we keep the Taylor series in the exponential, we don't expand the exponential itself as you seem to have done. Since the integrand sharply peaks close to $x=0$ most of the weight of the integral will come from a small region around $x=0$ so we can cut the Taylor series at second order which leads to
$$\int_0^{1000}e^{1000\frac{\sin(x)}{x}}\,{\rm d}x \sim \int_0^{1000} e^{1000 - \frac{1000}{6}x^{2}}\,{\rm d}x$$
where $\sim$ here means that the ratio of the LHS and RHS is close to $1$. Since the integrand drops very quickly to zero for large $x$ we can extend the integration region to $\infty$ without making much of an error (in absolute terms) and we are left with a standard gaussian integral which is easy to evaluate using the standard formula. This will give you something like
$$\int_0^{1000}e^{1000\frac{\sin(x)}{x}}\,{\rm d}x \sim e^{1000}\int_0^\infty e^{-\frac{1000}{6}x^2}\,{\rm d}x = e^{1000}\cdot C$$
for some constant $C$. To give a rough (order of magnitude) estimate of this first compute $a$ such that $e^{a} \approx 10$ (i.e. find an approximation for $\log(10)$) and write $e^{1000} \approx 10^{\frac{1000}{a}}$ and estimate the constant $C$. Note that you only need to use rough estimates like $\pi \approx 3$.