Integral of dx/2sinxcosx dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \cos{\left(x \right)} dx$ and substitute $0.5 du$:

      $$\int 0.5 u\, du$$

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

         $$\int u\, du = 0.5 \int u\, du$$

         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: $0.25 u^{2}$

      Now substitute $u$ back in:

      $$0.25 \sin^{2}{\left(x \right)}$$

   Method #2

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

      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- 0.5 du$:

      $$\int \left(- 0.5 u\right)\, du$$

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

         $$\int u\, du = - 0.5 \int u\, du$$

         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: $- 0.25 u^{2}$

      Now substitute $u$ back in:

      $$- 0.25 \cos^{2}{\left(x \right)}$$

2. Add the constant of integration:

   $$0.25 \sin^{2}{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$0.25 \sin^{2}{\left(x \right)}+ \mathrm{constant}$$