Catch me if you can!

We make use of complex numbers.

We know that

$e^{i \sin \theta } = \cos(\sin \theta) + i \sin ( \sin \theta)$

$\implies e^{\cos \theta} e^{i \sin \theta } = e^{\cos \theta} \cos(\sin \theta) + ie^{\cos \theta} \sin ( \sin \theta)$

$\implies e^{\cos \theta + i \sin \theta } = e^{\cos \theta} \cos(\sin \theta) + ie^{\cos \theta} \sin ( \sin \theta)$

Now, our integrand is the real part of $e^{\cos \theta + i \sin \theta }$ .

So our definite integral is nothing but

$\int\limits_0^\pi Re (e^{\cos \theta + i \sin \theta }) d \theta$

$= \int\limits_0^\pi Re( 1 + (\cos \theta + i \sin \theta) + \frac{(\cos \theta + i \sin \theta)^2}{2} + \cdot \cdot \cdot ) d \theta$

$= \int\limits_0^\pi Re( 1 + (\cos \theta + i \sin \theta) + \frac{\cos 2 \theta + i \sin 2 \theta}{2} + \cdot \cdot \cdot ) d \theta$

$= \int\limits_0^\pi ( 1 + \cos \theta + \frac{\cos 2 \theta }{2} + \cdot \cdot \cdot ) d \theta$

Since $\int\limits_0^\pi \cos n \theta d \theta = 0$ , where n is an integer, the value of our integral is nothing but $\pi$ .

So $a=b=1$ and $k=0$ . Then, $(a + b + k)^{12} = 2^{12} = 4096$ . Therefore, the sum of digits is $19$ .