Prove the improper integral is absolutely convergent $$\int_0^{\pi/2} \sin(\sec(x))dx$$
This is what I come up with:
Let $ \sec x=\frac{1}{t}, \cos x=\frac{1}{t} $, $x=\arccos\frac{1}{t}$, so
$dx = -\frac{1}{\sqrt{1-\frac{1}{t^2}}}\cdot \left(-\frac{1}{t^2}\right)dt$
$= \frac{1}{t\sqrt{t^2-1}}dt.$
Then when $x=0, t=1; x=\frac{\pi}{2}, t=+\infty $
 A: Even though $\sec x\to\infty$ as $x\to\pi/2$, this is not actually an improper integral, because $\sin(\sec x)$ is bounded in the finite interval $[0,\pi/2]$. (An integral is only improper if one or both of its limits is at infinity, or if the integrand is unbounded.) What we have here is
$$\left|\int_0^{\pi/2}\sin(\sec x)\,dx \right|\le\int_0^{\pi/2}|\sin(\sec x)|\,dx\le\int_0^{\pi/2}dx=\pi/2$$
You can, of course, change it into an improper integral with the substitution you came up with: Letting $\sec x=t$ (not $1/t$), we wind up with
$$\int_0^{\pi/2}\sin(\sec x)\,dx =\int_1^\infty{\sin t\over t\sqrt{t^2-1}}\,dt$$
and the latter is improper at both limits, $1$ (because of the $t^2-1$) and $\infty$.
A: Taking off from where Barry left off,we get the original integral as
$\int_1^\infty{\sin t\over t\sqrt{t^2-1}}dt$ which is clearly lesser than 
$\int_0^\infty{\sin t\over t\sqrt{t^2-1}}\,dt$ as sin(x)>0 in the interval (0,1)
now, 
$\int_0^\infty{\sin t\over t\sqrt{t^2-1}}\,dt$ =1/2$Re(\int_{-\infty}^\infty{{e^{it}}\over t\sqrt{t^2-1}}\,dt)$

now we take the following curve in the complex plane and apply Cauchy's theorem.
the integral over the semicircular arc vanishes and we are left with
$0=\oint_C f(z)\,dz=\int_{-R}^\epsilon f(x)\,dz-\int_{\gamma_\epsilon} f(z)\,dz+\int_\epsilon^R f(x)\,dx+\int_{\gamma_Rf(z)dz}$
now using the substitution
$\gamma(x)={\lim _{h\to 0}he^{ix}}.$
$\int_{\gamma_\epsilon}|{e^{ihe^{ix}}hie^{ix}\pi\over 2he^{ix}\sqrt{{(he^{ix})}^2-1}}\,| \,dz$=$\frac{h\pi i}{2hi}$
we find that integral equals to $\frac{\pi}{2}$
so we conclude that 
$\int_1^\infty{\sin t\over t\sqrt{t^2-1}}dt$<$\frac{\pi}{2}$
