A calculus problem by Agni Purani

Let $J = \frac{1}{\pi }\int_{0}^{2\pi}e^{\cos\theta }i\sin(\sin\theta) d\theta$

$I + J = \frac{1}{\pi }\int_{0}^{2\pi}e^{\cos\theta }(\cos(\sin\theta ) +i\sin(\sin\theta)) d\theta$

$\Rightarrow I + J = \frac{1}{\pi }\int_{0}^{2\pi}e^{\cos\theta }e^{i\sin\theta }$

$\Rightarrow I + J = \frac{1}{\pi }\int_{0}^{2\pi}e^{\cos\theta + i\sin\theta }$

$\Rightarrow I + J = \frac{1}{\pi }\int_{0}^{2\pi} \left ( 1 + (\cos\theta + i\sin\theta) + \frac{(\cos\theta + i\sin\theta)^{2}}{2!} + .... \right)$

and we now know that $I = Re( I + J)$

$\therefore I = \left [ \frac{1}{\pi } \left ( \theta + \sin\theta + \frac{\sin2\theta}{2 \times 2!} + .... \right) \right ]_{0}^{2\pi}$

$\Rightarrow I = \frac{1}{\pi }(2\pi) = \boxed{2}$