Proof that $J_{\nu}(x) \sim (x/2)^\nu / \Gamma(\nu+1) \; \text{as} \; \nu \rightarrow \infty$ I'm working through the exercises of Bender and Orszag's famous book, but I got stuck in 6.25 (a), in which it is asked to prove that
$$J_\nu (x) \sim (x/2)^\nu / \Gamma(\nu+1) \; \text{as} \; \nu \rightarrow \infty,$$
by using the following integral representation
$$J_\nu(x)=\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^\pi_0 \cos(x\cos\theta) \sin^{2\nu}\theta \, d\theta,$$
which is valid for $\nu > -1/2.$ ($J_\nu(x)$ is the $\nu$th-order Bessel function of the first kind.)
As the exercise belongs to section 6.4, which deals with Laplace's method and Watson's lemma, I thought I first had to perform a change of variables in order to get an integral of the form
$$I(x)=\int^b_a f(t)e^{x\phi(t)} \, dt.$$
So, I took $t=\cos\theta$ and obtained
$$\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^{1}_{-1} (1-t^2)^{p-\frac{1}{2}} e^{ixt} \, dt.$$
However, I cannot apply either Laplace's method or Watson's lemma, because the function $\phi$ I got is complex: $\phi(t)=it$.
What am I missing?
 A: To begin, rewrite the integral as
$$
\int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta.
$$
The quantity $\log \sin \theta$ has a maximum at $\theta = \pi/2$, and near there
$$
\log\sin\theta = -\frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2 + \cdots.
$$
Further
$$
\cos(x\cos\theta) = 1 + \cdots
$$
there, so by the Laplace method we have
$$
\int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta \sim \int_{-\infty}^{\infty} 1\cdot \exp\left[-2\nu \cdot \frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2\right]\,d\theta
$$
for large $\nu$.  Now simplify.
A: EDIT: 
As Antonio Vargas stated in the comments above, when $v$ is large, $\sin^{2v} \theta $ is small everywhere on the interval $[0, \pi]$ except $\theta = \frac{\pi}{2}$ (where $\sin (\theta) = 1$). 
So a slightly modified argument, noting that $\cos (x \cos \theta)=1$ when $\theta = \frac{\pi}{2}$, shows that it is also an asymptotic expansion of $J_{v}(x)$ as $v \to \infty$.
$ $
Both Wikipedia and Wolfram MathWorld state that is an asymptotic expansion as $x \to 0$.
http://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html (57)
$ $
Notice that for small $x$, $\cos (x \cos \theta)$ is essentially $1$.
Therefore, $$J_{v}(x) \sim \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi} \sin^{2v} \theta \ d \theta$$
$$ = 2 \ \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi /2} \sin^{2v} \theta \ d \theta$$
$$ = 2  \ \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi /2} \sin^{2(v+1/2)-1} (\theta) \cos^{2(1/2)-1} (\theta) \ d \theta$$
$$ = \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} B \left(v+\frac{1}{2},\frac{1}{2} \right)$$
$$ =\frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \frac{\Gamma(v+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(v+1)}$$
$$ = \frac{(\frac{x}{2})^{v}}{\Gamma(v+1)}$$
