another definite problem

= 2\pi

$Here $cos(sin\theta)=Re\left \{ cos(sin\theta)+isin(sin\theta)\right \}=Re(e^{isin(\theta)})$\\\\ So $\int_{0}^{2\pi}e^{cos\theta}\times Re(e^{isin\theta})d\theta=Re\left \{ \int_{0}^{2\pi}e^{cos\theta+isin\theta}d\theta \right\}$\\\\ $Re(\int_{0}^{2\pi}e^{e^{i\theta}}d\theta)=$

I had already done that but can't move forward...........

I = ∫e^cosθcos(sinθ)dθ = ∫Re(e^iθ)dθ

= ∫ {1 + cosθ1 + cos2θ2.1 + cos3θ3.2.1 + ....}dθ

taking limits from 0 to 2Î , only first term exists and hence I = 2Î