Integral of sin(2x)e^(cos(x)) dx

1. There are multiple ways to do this integral.

   Method #1

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

      $$\int 2 e^{\cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int e^{\cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx$$

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

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

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

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

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

               Then $\operatorname{du}{\left(u \right)} = 1$.

               To find $v{\left(u \right)}$:

               1. The integral of the exponential function is itself.

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

               Now evaluate the sub-integral.

            2. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$- e^{\cos{\left(x \right)}} \cos{\left(x \right)} + e^{\cos{\left(x \right)}}$$

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

   Method #2

   1. Rewrite the integrand:

      $$e^{\cos{\left(x \right)}} \sin{\left(2 x \right)} = 2 e^{\cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}$$

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

      $$\int 2 e^{\cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int e^{\cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx$$

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

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

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

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

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

               Then $\operatorname{du}{\left(u \right)} = 1$.

               To find $v{\left(u \right)}$:

               1. The integral of the exponential function is itself.

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

               Now evaluate the sub-integral.

            2. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$- e^{\cos{\left(x \right)}} \cos{\left(x \right)} + e^{\cos{\left(x \right)}}$$

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

2. Now simplify:

   $$2 \left(1 - \cos{\left(x \right)}\right) e^{\cos{\left(x \right)}}$$

3. Add the constant of integration:

   $$2 \left(1 - \cos{\left(x \right)}\right) e^{\cos{\left(x \right)}}+ \mathrm{constant}$$

The answer is:

$$2 \left(1 - \cos{\left(x \right)}\right) e^{\cos{\left(x \right)}}+ \mathrm{constant}$$