Integral of sinx*cos^6x*dx dx

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

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

   $$\int u^{6}\, du$$

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

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

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

         $$\int u^{6}\, du = \frac{u^{7}}{7}$$

      So, the result is: $- \frac{u^{7}}{7}$

   Now substitute $u$ back in:

   $$- \frac{\cos^{7}{\left(x \right)}}{7}$$

2. Add the constant of integration:

   $$- \frac{\cos^{7}{\left(x \right)}}{7}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos^{7}{\left(x \right)}}{7}+ \mathrm{constant}$$