Integral of Sin^5*x/2*cos*x/2 dx

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

   $$\int \frac{\sin^{5}{\left(x \right)} \cos{\left(x \right)}}{2 \cdot 2}\, dx = \frac{\int \sin^{5}{\left(x \right)} \cos{\left(x \right)}\, dx}{4}$$

   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 $du$:

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

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

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

         Now substitute $u$ back in:

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

      Method #2

      1. Rewrite the integrand:

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

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

         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:

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

            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:

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

   So, the result is: $\frac{\sin^{6}{\left(x \right)}}{24}$

2. Add the constant of integration:

   $$\frac{\sin^{6}{\left(x \right)}}{24}+ \mathrm{constant}$$

The answer is:

$$\frac{\sin^{6}{\left(x \right)}}{24}+ \mathrm{constant}$$