Integral of sinxcos(cosx-1) 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 $- du$:

      $$\int \cos{\left(u \right)}\, du$$

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

         $$\int \left(- \cos{\left(u \right)}\right)\, du = - \int \cos{\left(u \right)}\, du$$

         1. The integral of cosine is sine:

            $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

      $$\sin{\left(x \right)} \cos{\left(\cos{\left(x \right)} - 1 \right)} = \sin{\left(1 \right)} \sin{\left(x \right)} \sin{\left(\cos{\left(x \right)} \right)} + \sin{\left(x \right)} \cos{\left(1 \right)} \cos{\left(\cos{\left(x \right)} \right)}$$

   2. Integrate term-by-term:

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

         $$\int \sin{\left(1 \right)} \sin{\left(x \right)} \sin{\left(\cos{\left(x \right)} \right)}\, dx = \sin{\left(1 \right)} \int \sin{\left(x \right)} \sin{\left(\cos{\left(x \right)} \right)}\, dx$$

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

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

            $$\int \sin{\left(u \right)}\, du$$

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

               $$\int \left(- \sin{\left(u \right)}\right)\, du = - \int \sin{\left(u \right)}\, du$$

               1. The integral of sine is negative cosine:

                  $$\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$$

               So, the result is: $\cos{\left(u \right)}$

            Now substitute $u$ back in:

            $$\cos{\left(\cos{\left(x \right)} \right)}$$

         So, the result is: $\sin{\left(1 \right)} \cos{\left(\cos{\left(x \right)} \right)}$

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

         $$\int \sin{\left(x \right)} \cos{\left(1 \right)} \cos{\left(\cos{\left(x \right)} \right)}\, dx = \cos{\left(1 \right)} \int \sin{\left(x \right)} \cos{\left(\cos{\left(x \right)} \right)}\, dx$$

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

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

            $$\int \cos{\left(u \right)}\, du$$

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

               $$\int \left(- \cos{\left(u \right)}\right)\, du = - \int \cos{\left(u \right)}\, du$$

               1. The integral of cosine is sine:

                  $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

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

            Now substitute $u$ back in:

            $$- \sin{\left(\cos{\left(x \right)} \right)}$$

         So, the result is: $- \sin{\left(\cos{\left(x \right)} \right)} \cos{\left(1 \right)}$

      The result is: $- \sin{\left(\cos{\left(x \right)} \right)} \cos{\left(1 \right)} + \sin{\left(1 \right)} \cos{\left(\cos{\left(x \right)} \right)}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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