Calculate $ \displaystyle \left|\frac {\int_0^{\frac {\pi}{2}} (x\cos x + 1)e^{\sin x}\ dx}{\int_0^{\frac {\pi}{2}} (x\sin x - 1)e^{\cos x}\ dx}\right|$.
1. We start by evaluating the numerator of the given expression:
   \[
   \int_0^{\frac{\pi}{2}} (x \cos x + 1)e^{\sin x} \, dx
   \]
   We can split this integral into two parts:
   \[
   \int_0^{\frac{\pi}{2}} x \cos x \cdot e^{\sin x} \, dx + \int_0^{\frac{\pi}{2}} e^{\sin x} \, dx
   \]

2. For the first part, we use integration by parts. Let \( u = x \) and \( dv = \cos x \cdot e^{\sin x} \, dx \). Then, \( du = dx \) and we need to find \( v \):
   \[
   v = \int \cos x \cdot e^{\sin x} \, dx
   \]
   Let \( t = \sin x \), then \( dt = \cos x \, dx \). The integral becomes:
   \[
   v = \int e^t \, dt = e^t = e^{\sin x}
   \]
   Now, applying integration by parts:
   \[
   \int_0^{\frac{\pi}{2}} x \cos x \cdot e^{\sin x} \, dx = \left[ x e^{\sin x} \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} e^{\sin x} \, dx
   \]
   Evaluating the boundary term:
   \[
   \left[ x e^{\sin x} \right]_0^{\frac{\pi}{2}} = \left( \frac{\pi}{2} e^1 \right) - \left( 0 \cdot e^0 \right) = \frac{\pi}{2} e
   \]
   Therefore:
   \[
   \int_0^{\frac{\pi}{2}} x \cos x \cdot e^{\sin x} \, dx = \frac{\pi}{2} e - \int_0^{\frac{\pi}{2}} e^{\sin x} \, dx
   \]

3. Adding the second part of the integral:
   \[
   \int_0^{\frac{\pi}{2}} (x \cos x + 1)e^{\sin x} \, dx = \left( \frac{\pi}{2} e - \int_0^{\frac{\pi}{2}} e^{\sin x} \, dx \right) + \int_0^{\frac{\pi}{2}} e^{\sin x} \, dx = \frac{\pi}{2} e
   \]

4. Next, we evaluate the denominator of the given expression:
   \[
   \int_0^{\frac{\pi}{2}} (x \sin x - 1)e^{\cos x} \, dx
   \]
   We can split this integral into two parts:
   \[
   \int_0^{\frac{\pi}{2}} x \sin x \cdot e^{\cos x} \, dx - \int_0^{\frac{\pi}{2}} e^{\cos x} \, dx
   \]

5. For the first part, we use integration by parts. Let \( u = x \) and \( dv = \sin x \cdot e^{\cos x} \, dx \). Then, \( du = dx \) and we need to find \( v \):
   \[
   v = \int \sin x \cdot e^{\cos x} \, dx
   \]
   Let \( t = \cos x \), then \( dt = -\sin x \, dx \). The integral becomes:
   \[
   v = -\int e^t \, dt = -e^t = -e^{\cos x}
   \]
   Now, applying integration by parts:
   \[
   \int_0^{\frac{\pi}{2}} x \sin x \cdot e^{\cos x} \, dx = \left[ -x e^{\cos x} \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} e^{\cos x} \, dx
   \]
   Evaluating the boundary term:
   \[
   \left[ -x e^{\cos x} \right]_0^{\frac{\pi}{2}} = \left( -\frac{\pi}{2} e^0 \right) - \left( 0 \cdot e^1 \right) = -\frac{\pi}{2}
   \]
   Therefore:
   \[
   \int_0^{\frac{\pi}{2}} x \sin x \cdot e^{\cos x} \, dx = -\frac{\pi}{2} + \int_0^{\frac{\pi}{2}} e^{\cos x} \, dx
   \]

6. Subtracting the second part of the integral:
   \[
   \int_0^{\frac{\pi}{2}} (x \sin x - 1)e^{\cos x} \, dx = \left( -\frac{\pi}{2} + \int_0^{\frac{\pi}{2}} e^{\cos x} \, dx \right) - \int_0^{\frac{\pi}{2}} e^{\cos x} \, dx = -\frac{\pi}{2}
   \]

7. Finally, we compute the given expression:
   \[
   \left| \frac{\int_0^{\frac{\pi}{2}} (x \cos x + 1)e^{\sin x} \, dx}{\int_0^{\frac{\pi}{2}} (x \sin x - 1)e^{\cos x} \, dx} \right| = \left| \frac{\frac{\pi}{2} e}{-\frac{\pi}{2}} \right| = \left| -e \right| = e
   \]

The final answer is \( \boxed{ e } \)