Integral $\int_0^\frac{\pi}{2}\sin(\csc{x})\ \mathrm{d}x$ $$\int_0^\frac{\pi}{2}\sin(\csc{x})\ \mathrm{d}x.$$ 
Wolfram Alpha gives the value of this as 1.00071. Any help would be greatly appreciated. Thank you in advance!
 A: Let's make the substitution $t=\frac{1}{\sin x} $, we have 
$$x = \arcsin{\frac{1}{t}} $$
and
$$ \mathrm{d} x = -\frac{\mathrm{d}t}{t\sqrt{t^2-1}}.  $$
The integral becomes
$$\int_1^{+\infty }\frac{\sin{t}}{t\sqrt{t^2-1}}\mathrm{d}t.$$
We notice that the integral converges:
$$\Bigg|\int_1^{+\infty }\frac{\sin{t}}{t\sqrt{t^2-1}}\mathrm{d}t\Bigg| \le \int_1^{+\infty }\Bigg|\frac{\sin{t}}{t\sqrt{t^2-1}}\Bigg|\mathrm{d}t \le \int_1^{+\infty }\frac{1}{t\sqrt{t^2-1}}\mathrm{d}t = \frac{\pi}{2}$$
(just make the sostitution $u = \sqrt{t^2-1} $ in the last integral).
Let's define now
$$I(\alpha)=\int_1^{+\infty }\frac{\sin{(\alpha t)}}{t\sqrt{t^2-1}}\mathrm{d}t.$$
We want to calculate $I(1)$ and we know that $I(0)= 0$. So differentiating:
$$I'(\alpha)= \int_1^{+\infty }\frac{\cos{(\alpha t)}}{\sqrt{t^2-1}}\mathrm{d}t = -\frac{\pi}{2}Y_0(\alpha)$$
where $Y_0(\alpha)$ is the Bessel function of second Kind (see formula 9 here)
So the problem is to evaluate $I(1)$ where
$$ 
\begin{cases}
I'(\alpha)=-\frac{\pi}{2}Y_0(\alpha)\\[1ex]
I(0)= 0
\end{cases}
$$
From here we know that
$$
I(\alpha)=-\int Y_0(\alpha) \mathrm{d}\alpha = -\frac{\pi^2}{4} \alpha \bigg(Y_0(\alpha)H_{-1}(\alpha)-Y_1(\alpha)H_0(\alpha)\bigg) + C
$$
where $H_z(\alpha)$ is the Struve function (here for more details).
So if we impose the condition $I(0)=0$ we have
$$
I(0)= C = 0
$$
and the results is
$$
I(1)=-\frac{\pi^2}{4}  \Big[ Y_0(1)H_{-1}(1)+Y_1(1)H_0(1)) \Big]
$$
The answer is not pretty, but with some calculation we have that
$$\int_{0}^{\frac{\pi}{2}} \sin{\csc{x}} \ \mathrm{d}x \approx 1.0007$$
A: With CAS help and Mellin Transform:
$$\int_0^{\frac{\pi }{2}} \sin (\csc (x)) \, dx=\\\mathcal{M}_s^{-1}\left[\int_0^{\frac{\pi }{2}} \mathcal{M}_a[\sin (a \csc (x))](s) \, dx\right](1)=\\\mathcal{M}_s^{-1}\left[\int_0^{\frac{\pi }{2}} \csc ^{-s}(x) \Gamma
   (s) \sin \left(\frac{\pi  s}{2}\right) \, dx\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{\sqrt{\pi } \Gamma (s) \Gamma \left(\frac{1+s}{2}\right) \sin \left(\frac{\pi  s}{2}\right)}{2 \Gamma
   \left(1+\frac{s}{2}\right)}\right](1)=\\\frac{1}{2} \pi  G_{1,3}^{2,0}\left(\frac{1}{2},\frac{1}{2}|
\begin{array}{c}
 1 \\
 \frac{1}{2},\frac{1}{2},0 \\
\end{array}
\right)\approx 1.00070623668845484$$
where: $\frac{1}{2} \pi  G_{1,3}^{2,0}\left(\frac{1}{2},\frac{1}{2}|
\begin{array}{c}
 1 \\
 \frac{1}{2},\frac{1}{2},0 \\
\end{array}
\right)$ is the Meijer G function.
Mathematica code:
HoldForm[Integrate[Sin[Csc[x]], {x, 0, Pi/2}] == 
   InverseMellinTransform[
    Integrate[MellinTransform[Sin[a Csc[x]], a, s], {x, 0, Pi/2}], s, 
    1] == InverseMellinTransform[
    Integrate[Csc[x]^-s Gamma[s] Sin[(\[Pi] s)/2], {x, 0, Pi/2}], s, 
    1] == InverseMellinTransform[(
    Sqrt[\[Pi]] Gamma[s] Gamma[(1 + s)/2] Sin[(\[Pi] s)/2])/(
    2 Gamma[1 + s/2]), s, 1] == 
   1/2 \[Pi] MeijerG[{{}, {1}}, {{1/2, 1/2}, {0}}, 1/2, 1/
     2] \[TildeTilde] 1.000706236688454839] // TeXForm
