Let's consider the integrer:

$\displaystyle \int -cos(2x)e^{cos(x)+sin(x)} dx$

Let's assume that:

$\displaystyle cos(x)+sin(x) = u$

$\displaystyle \Big (cos(x)-sin(x) \Big ) dx = du$

$\displaystyle \Bigg ( \Big (cos(x)-sin(x) \Big ) \frac{cos(x)+sin(x)}{cos(x)+sin(x)} \Bigg )dx = du$

$\displaystyle \frac{cos^{2}(x)-sin^{2}(x)}{cos(x)+sin(x)} dx = du$

$\displaystyle \frac{cos(2x)}{u} dx = du$

$\displaystyle cos(2x) dx = u du$

then our interger:

$\displaystyle - \int ue^udu$

next:

$\displaystyle f(x) = e^u, f'(x) = e^u$

oraz

$\displaystyle g(x) = u, g'(x) = 1$

and finally we get:

$\displaystyle - \int ue^udu = ue^u - \int e^u = ue^u - e^u +C = e^u(u - 1) + C$

so:

$\displaystyle - \int cos(2x)e^{cos(x)+sin(x)} = e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big ) + C$

...and I am not sure if that solution is correct.