Integral of sinxcos^5x dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int u^{5}\, du$$

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

         $$\int \left(- u^{5}\right)\, du = - \int u^{5}\, du$$

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

            $$\int u^{5}\, du = \frac{u^{6}}{6}$$

         So, the result is: $- \frac{u^{6}}{6}$

      Now substitute $u$ back in:

      $$- \frac{\cos^{6}{\left(x \right)}}{6}$$

   Method #2

   1. Rewrite the integrand:

      $$\sin{\left(x \right)} \cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos{\left(x \right)}$$

   2. Let $u = 1 - \sin^{2}{\left(x \right)}$.

      Then let $du = - 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- \frac{du}{2}$:

      $$\int \frac{u^{2}}{4}\, du$$

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

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

         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: $- \frac{u^{3}}{6}$

      Now substitute $u$ back in:

      $$- \frac{\left(1 - \sin^{2}{\left(x \right)}\right)^{3}}{6}$$

2. Add the constant of integration:

   $$- \frac{\cos^{6}{\left(x \right)}}{6}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos^{6}{\left(x \right)}}{6}+ \mathrm{constant}$$