Integral of -4*sin(t)*cos(t) dt

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \cos{\left(t \right)} dt$ and substitute $- 4 du$:

      $$\int \left(- 4 u\right)\, du$$

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

         $$\int u\, du = - 4 \int u\, du$$

         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: $- 2 u^{2}$

      Now substitute $u$ back in:

      $$- 2 \sin^{2}{\left(t \right)}$$

   Method #2

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

      Then let $du = - \sin{\left(t \right)} dt$ and substitute $4 du$:

      $$\int 4 u\, du$$

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

         $$\int u\, du = 4 \int u\, du$$

         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: $2 u^{2}$

      Now substitute $u$ back in:

      $$2 \cos^{2}{\left(t \right)}$$

2. Add the constant of integration:

   $$- 2 \sin^{2}{\left(t \right)}+ \mathrm{constant}$$

The answer is:

$$- 2 \sin^{2}{\left(t \right)}+ \mathrm{constant}$$