Integral of 1/2sin(2x)*exp(sin(x)) dx

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

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

   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:

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

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

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

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

            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.

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

            Now substitute $u$ back in:

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

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

      Method #2

      1. Rewrite the integrand:

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

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

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

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

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

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

            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.

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

            Now substitute $u$ back in:

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

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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