Integral of 6cos^2x*sinx dx

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

   Then let $du = - \sin{\left(x \right)} dx$ and substitute $- 6 du$:

   $$\int \left(- 6 u^{2}\right)\, du$$

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

      $$\int u^{2}\, du = - 6 \int u^{2}\, du$$

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

         $$\int u^{2}\, du = \frac{u^{3}}{3}$$

      So, the result is: $- 2 u^{3}$

   Now substitute $u$ back in:

   $$- 2 \cos^{3}{\left(x \right)}$$

2. Add the constant of integration:

   $$- 2 \cos^{3}{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- 2 \cos^{3}{\left(x \right)}+ \mathrm{constant}$$