Integral of (e^(3cosx+2))sinx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 \cos{\left(x \right)} + 2$.

      Then let $du = - 3 \sin{\left(x \right)} dx$ and substitute $- \frac{du}{3}$:

      $$\int \left(- \frac{e^{u}}{3}\right)\, du$$

      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: $- \frac{e^{u}}{3}$

      Now substitute $u$ back in:

      $$- \frac{e^{3 \cos{\left(x \right)} + 2}}{3}$$

   Method #2

   1. Rewrite the integrand:

      $$e^{3 \cos{\left(x \right)} + 2} \sin{\left(x \right)} = e^{2} e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}$$

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

      $$\int e^{2} e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}\, dx = e^{2} \int e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}\, dx$$

      1. Let $u = 3 \cos{\left(x \right)}$.

         Then let $du = - 3 \sin{\left(x \right)} dx$ and substitute $- \frac{du}{3}$:

         $$\int \left(- \frac{e^{u}}{3}\right)\, du$$

         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: $- \frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$- \frac{e^{3 \cos{\left(x \right)}}}{3}$$

      So, the result is: $- \frac{e^{2} e^{3 \cos{\left(x \right)}}}{3}$

   Method #3

   1. Rewrite the integrand:

      $$e^{3 \cos{\left(x \right)} + 2} \sin{\left(x \right)} = e^{2} e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}$$

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

      $$\int e^{2} e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}\, dx = e^{2} \int e^{3 \cos{\left(x \right)}} \sin{\left(x \right)}\, dx$$

      1. Let $u = 3 \cos{\left(x \right)}$.

         Then let $du = - 3 \sin{\left(x \right)} dx$ and substitute $- \frac{du}{3}$:

         $$\int \left(- \frac{e^{u}}{3}\right)\, du$$

         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: $- \frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$- \frac{e^{3 \cos{\left(x \right)}}}{3}$$

      So, the result is: $- \frac{e^{2} e^{3 \cos{\left(x \right)}}}{3}$

2. Now simplify:

   $$- \frac{e^{3 \cos{\left(x \right)} + 2}}{3}$$

3. Add the constant of integration:

   $$- \frac{e^{3 \cos{\left(x \right)} + 2}}{3}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{3 \cos{\left(x \right)} + 2}}{3}+ \mathrm{constant}$$