Integral of sinxe^(cosx-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

   Method #2

   1. Rewrite the integrand:

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

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

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

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

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

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

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

   Method #3

   1. Rewrite the integrand:

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

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

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

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

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

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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