Integral of -2e^(-y)cosx dy

1. The integral of a constant times a function is the constant times the integral of the function:

   $$\int - 2 e^{- y} \cos{\left(x \right)}\, dy = \cos{\left(x \right)} \int \left(- 2 e^{- y}\right)\, dy$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- 2 e^{- y}\right)\, dy = - 2 \int e^{- y}\, dy$$

      1. Let $u = - y$.

         Then let $du = - dy$ and substitute $- du$:

         $$\int \left(- e^{u}\right)\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\text{False}$$

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $- e^{u}$

         Now substitute $u$ back in:

         $$- e^{- y}$$

      So, the result is: $2 e^{- y}$

   So, the result is: $2 e^{- y} \cos{\left(x \right)}$

2. Add the constant of integration:

   $$2 e^{- y} \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$2 e^{- y} \cos{\left(x \right)}+ \mathrm{constant}$$