Integral of sinx(1+cosx) 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 $- du$:

      $$\int \left(- u\right)\, du$$

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

         $$\int u\, du = - \int u\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u\, du = \frac{u^{2}}{2}$$

         So, the result is: $- \frac{u^{2}}{2}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         $$\int \left(- u\right)\, du$$

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

            $$\int u\, du = - \int u\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u\, du = \frac{u^{2}}{2}$$

            So, the result is: $- \frac{u^{2}}{2}$

         Now substitute $u$ back in:

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

      1. The integral of sine is negative cosine:

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

      The result is: $- \frac{\cos^{2}{\left(x \right)}}{2} - \cos{\left(x \right)}$

   Method #3

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         $$\int \left(- u\right)\, du$$

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

            $$\int u\, du = - \int u\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u\, du = \frac{u^{2}}{2}$$

            So, the result is: $- \frac{u^{2}}{2}$

         Now substitute $u$ back in:

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

      1. The integral of sine is negative cosine:

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

      The result is: $- \frac{\cos^{2}{\left(x \right)}}{2} - \cos{\left(x \right)}$

2. Add the constant of integration:

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

The answer is:

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