Interesting real integrals from $ I=\oint\limits_{C}\sin(\frac{1}{z})\,\mathrm{dz} = 2\pi i $ I have separated and equated Real and Imaginary Parts of the following equation by plugging $\mathrm{z}=e^{it}$    :
$$I=\oint\limits_{C}\sin\frac{1}{z}\,\mathrm{dz} = 2\pi i,$$
where $C$ is the boundary of unit circle centered at origin.
which gives $$ K=\int_{-\pi}^{+\pi}e^{it}\sin(e^{-it})\mathrm{dt} = 2\pi$$
simplifying gives the following:-
$\int_{0}^{\frac{\pi}{2}}\cos(t)\sin(\cos(t))\cosh(\sin(t))+\sin(t)\cos(\cos(t))\sinh(\sin(t))\mathrm{dt} = \frac{\pi}{2}$
the first and second part of the above integral translates to $I_1$ and $I_2$ respectively under a simple substitution for each.
$$ I_1 = \int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\mathrm{dx}\\ I_2 = \int_{0}^1 \cos(x)\sinh(\sqrt{1-x^2})\mathrm{dx}$$
which implies, $I_1 + I_2 = \frac{\Large\pi}{2}$
Interestingly, I have found that $I_1 $ and $ I_2$ are equal (using a calculator), but stuck proving them.
Any ideas how they are equal or finding another method to calculate $I_1 $ or $ I_2$?
EXTRA: some integrals that show up but eventually cancel out( as they are odd functions, $f(x)=f(-x)$ ) in calculating Imaginary part of equation $K$:-
$$ I_3 = \int_{0}^1 \sin(x)\cosh(\sqrt{1-x^2})\mathrm{dx}\\ I_4 = \int_{0}^1 \sinh(x)\cos(\sqrt{1-x^2})\mathrm{dx}$$
however $I_3 $ and $ I_4$ are not equal and their numerical values are approx. 0.584 and 0.418 , can $I_3 $ and $ I_4$ be calculated in closed form?
 A: Partial answer:
$$\begin{align}
I_1&=\int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\,dx\\
&\stackrel{x\to\sqrt{1-t^2}}{=}
\int_{0}^1 \cosh(\sqrt{1-t^2})\sin(t)\frac{t\,dt}{\sqrt{1-t^2}}\\
&=-\int_{0}^1 \sin(t)\,d(\sinh(\sqrt{1-t^2}))\\
&=\left[-\sin(t)\sinh(\sqrt{1-t^2})\right]_0^1+\int_{0}^1 \sinh(\sqrt{1-t^2})\,d (\sin(t))\\
&=\int_{0}^1 \sinh(\sqrt{1-t^2})\cos(t)\, dt=I_2.
\end{align}$$
It follows $I_1=I_2=\frac\pi4$.
Applying the same method one can show $I_3+I_4=\cosh(1)-\cos(1)$.
A: The value of the integral $$I_{3} = \int_{0}^{1} \sin(x) \cosh \left( \sqrt{1-x^{2}} \right) \, \mathrm dx $$ can  be expressed in terms of the cosine integral $\operatorname{Ci}(x)$ and the hyperbolic cosine integral $\operatorname{Chi}(x)$.
$$ \begin{align} I_{3} &= \int_{0}^{\pi/2} \sin(\sin t) \cosh(\cos t) \cos(t) \, \mathrm dt \\ &= \Im \int_{0}^{\pi/2} \sinh(e^{it}) \cos (t) \, \mathrm dt \\ &= \Im \int_{C} \sinh(z) \,  \frac{z+\frac{1}{z}}{2} \frac{dz}{iz} \\ &= -\frac{1}{2} \, \Re \int_{C} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz, \end{align}$$
where $C$ is the portion of the unit circle in the first quadrant of the complex plane.
But since the integrand is analytic in the first quadrant, we have$$ \begin{align} I_{3} &= - \frac{1}{2} \, \Re \int_{1}^{i} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} \Bigg|^{i}_{1} + \int_{1}^{i} \frac{\cosh (z)}{z} \, \mathrm dz \right) \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} + \operatorname{Chi}(z) \Bigg|_{1}^{i} \right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \Re \left(\operatorname{Chi}(i)  \right)- \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \operatorname{Ci}(1) - \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= \frac{1}{2} \left(\sin(1) - \cos(1) - \operatorname{Ci}(1) + \frac{1}{e} + \operatorname{Chi}(1) \right) \\ & \approx 0.58475. \end{align}$$

The evaluation of  $$I_{4} = \int_{0}^{\pi/2} \sinh(\sin t) \cos(\cos t) \cos (t) \, \mathrm dt =  \Im \int_{0}^{\pi/2} \sin(e^{it}) \cos(t) \, \mathrm dt$$ should be similar.
