How to evaluate the integrals: $\int_0^{\pi/2}(\cos^2 \frac{x}{2}+x\cos x)e^{\sin x}dx$ $$I_1=\int_0^{\pi/4}\frac{\sin x}{x\cos^2 x}dx$$
With this integral, I can't solve. I think setting $x=\frac{\pi}{4}-t$ is ok. But it seeems to be wrong.
$$I_3=\int_0^{\pi/2}(\cos^2 \frac{x}{2}+x\cos x)e^{\sin x}dx$$
I'd tried to split this integral, and then using method similar $I_1$ . But I can't get the result :( . And I need your help! Thanks. 
 A: This is a partial answer as I only solve $I_3$, not $I_1$ yet.
$$
I_3=\int_0^{\pi/2} \cos^2\big(\frac{x}{2}\big) e^{\sin x} dx +\int_0^{\pi/2} x\cos x e^{\sin x}dx.
$$
We can use a trig identity $2\cos^2(x/2)=(1+\cos x)$ on the first integral to obtain
$$
\int_0^{\pi/2} \cos^2\big(\frac{x}{2}\big) e^{\sin x} dx=\frac{1}{2}\int_0^{\pi/2} e^{\sin x}dx +\frac{1}{2}\int_0^{\pi/2}\cos x e^{\ sin x} dx=\frac{\pi}{4} \big(L_o(1)+I_0(1)\big)+\frac{1}{2}(e-1)
$$
where the first integral is given by the Struvel function L, and the modified Bessel function of the first kind I.  The second integral is trivially done by using $u=\sin x$.   We can now re-write $I_3$ as
$$
I_3=\frac{\pi}{4} \big(L_o(1)+I_0(1)\big)+\frac{1}{2}(e-1)+\int_0^{\pi/2} x\cos x e^{\sin x}dx.$$
This last integral can also be expressed in terms of Struvel and modified Bessel functions, but we first use partial integration to notice this, $u=x, du=dx, dv=\cos x \exp(\sin x)dx,v= \exp(\sin x)$, thus we obtain
$$
\int_0^{\pi/2} x\cos x e^{\sin x}dx=\frac{\pi e}{2}-\int_0^{\pi/2} e^{\sin x}dx= \frac{\pi}{2} \big(e-L_0(1)-I_0(1)\big),
$$
Thus we can simplify $I_3$ and write
$$
I_3=\frac{1}{4} \big(2e(1+\pi)-\pi (L_0(1)+I_0(1))-2\big)
$$
Note: Struvel L function http://functions.wolfram.com/Bessel-TypeFunctions/StruveL/.  The modified bessel function $I_0$ is given by 
$$
I_0(z)=\frac{1}{\pi}\int_0^\pi e^{z\cos x}dx, \to \quad 
I_0(1)=\frac{1}{\pi}\int_0^\pi e^{\cos x}dx.
$$
$$
L_0(1)=\frac{2}{\pi} \int_0^{\pi/2} e^{\sin x} dx-I_0(1)=\frac{2}{\pi} \int_0^{\pi/2} e^{\sin x} dx-\frac{1}{\pi}\int_0^\pi e^{\cos x}dx.
$$
