【答案】\(\boldsymbol{\frac{4}{3} - 2\sin1}\) 

【解析】由题易知可作曲线如右图所示. 

记\( L_0 \)是从\( x = -1 \)到\( x = 1 \)的直线， 

并记曲线积\( I = \int_{L} (y + \cos x)dx + (2x + \cos y)dy \) 

则在\( L_0 \)与\( L \)所围的封闭区域可用格林公式 

即\( I_1 = \oint_{L_0 + L} (y + \cos x)dx + (2x + \cos y)dy \) 

\( = \iint_{D} 2 - 1d\sigma = \int_{-1}^{1} dx\int_{0}^{1 - x^2} dy = \int_{-1}^{1} (1 - x^2)dx = \frac{4}{3} \) 

又\( I_2 = \int_{L_0} (y + \cos x)dx + (2x + \cos y)dy = \int_{-1}^{1} \cos xdx = 2\sin1 \)，故\( I = \frac{4}{3} - 2\sin1 \)