Integral of cot^5x dx

1. Rewrite the integrand:

   $$\cot^{5}{\left(x \right)} = \left(\csc^{2}{\left(x \right)} - 1\right)^{2} \cot{\left(x \right)}$$

2. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \frac{- u^{2} + 2 u - 1}{4 u}\, du$$

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

         $$\int \frac{- u^{2} + 2 u - 1}{2 u}\, du = \frac{\int \frac{- u^{2} + 2 u - 1}{u}\, du}{2}$$

         1. Rewrite the integrand:

            $$\frac{- u^{2} + 2 u - 1}{u} = - u + 2 - \frac{1}{u}$$

         2. Integrate term-by-term:

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

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

               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: $- \frac{u^{2}}{2}$

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 2\, du = 2 u$$

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

               $$\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$$

               1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

               So, the result is: $- \log{\left(u \right)}$

            The result is: $- \frac{u^{2}}{2} + 2 u - \log{\left(u \right)}$

         So, the result is: $- \frac{u^{2}}{4} + u - \frac{\log{\left(u \right)}}{2}$

      Now substitute $u$ back in:

      $$- \frac{\log{\left(\csc^{2}{\left(x \right)} \right)}}{2} - \frac{\csc^{4}{\left(x \right)}}{4} + \csc^{2}{\left(x \right)}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(\csc^{2}{\left(x \right)} - 1\right)^{2} \cot{\left(x \right)} = \cot{\left(x \right)} \csc^{4}{\left(x \right)} - 2 \cot{\left(x \right)} \csc^{2}{\left(x \right)} + \cot{\left(x \right)}$$

   2. Integrate term-by-term:

      1. Let $u = \csc^{4}{\left(x \right)}$.

         Then let $du = - 4 \cot{\left(x \right)} \csc^{4}{\left(x \right)} dx$ and substitute $- \frac{du}{4}$:

         $$\int \frac{1}{16}\, du$$

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

            $$\int \left(- \frac{1}{4}\right)\, du = - \frac{\int 1\, du}{4}$$

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            So, the result is: $- \frac{u}{4}$

         Now substitute $u$ back in:

         $$- \frac{\csc^{4}{\left(x \right)}}{4}$$

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

         $$\int \left(- 2 \cot{\left(x \right)} \csc^{2}{\left(x \right)}\right)\, dx = - 2 \int \cot{\left(x \right)} \csc^{2}{\left(x \right)}\, dx$$

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

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

            $$\int \frac{1}{4}\, du$$

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

               $$\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$$

               1. The integral of a constant is the constant times the variable of integration:

                  $$\int 1\, du = u$$

               So, the result is: $- \frac{u}{2}$

            Now substitute $u$ back in:

            $$- \frac{\csc^{2}{\left(x \right)}}{2}$$

         So, the result is: $\csc^{2}{\left(x \right)}$

      1. Rewrite the integrand:

         $$\cot{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$$

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

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

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(\sin{\left(x \right)} \right)}$$

      The result is: $\log{\left(\sin{\left(x \right)} \right)} - \frac{\csc^{4}{\left(x \right)}}{4} + \csc^{2}{\left(x \right)}$

   Method #3

   1. Rewrite the integrand:

      $$\left(\csc^{2}{\left(x \right)} - 1\right)^{2} \cot{\left(x \right)} = \cot{\left(x \right)} \csc^{4}{\left(x \right)} - 2 \cot{\left(x \right)} \csc^{2}{\left(x \right)} + \cot{\left(x \right)}$$

   2. Integrate term-by-term:

      1. Let $u = \csc^{4}{\left(x \right)}$.

         Then let $du = - 4 \cot{\left(x \right)} \csc^{4}{\left(x \right)} dx$ and substitute $- \frac{du}{4}$:

         $$\int \frac{1}{16}\, du$$

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

            $$\int \left(- \frac{1}{4}\right)\, du = - \frac{\int 1\, du}{4}$$

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            So, the result is: $- \frac{u}{4}$

         Now substitute $u$ back in:

         $$- \frac{\csc^{4}{\left(x \right)}}{4}$$

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

         $$\int \left(- 2 \cot{\left(x \right)} \csc^{2}{\left(x \right)}\right)\, dx = - 2 \int \cot{\left(x \right)} \csc^{2}{\left(x \right)}\, dx$$

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

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

            $$\int \frac{1}{4}\, du$$

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

               $$\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$$

               1. The integral of a constant is the constant times the variable of integration:

                  $$\int 1\, du = u$$

               So, the result is: $- \frac{u}{2}$

            Now substitute $u$ back in:

            $$- \frac{\csc^{2}{\left(x \right)}}{2}$$

         So, the result is: $\csc^{2}{\left(x \right)}$

      1. Rewrite the integrand:

         $$\cot{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$$

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

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

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(\sin{\left(x \right)} \right)}$$

      The result is: $\log{\left(\sin{\left(x \right)} \right)} - \frac{\csc^{4}{\left(x \right)}}{4} + \csc^{2}{\left(x \right)}$

3. Add the constant of integration:

   $$- \frac{\log{\left(\csc^{2}{\left(x \right)} \right)}}{2} - \frac{\csc^{4}{\left(x \right)}}{4} + \csc^{2}{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- \frac{\log{\left(\csc^{2}{\left(x \right)} \right)}}{2} - \frac{\csc^{4}{\left(x \right)}}{4} + \csc^{2}{\left(x \right)}+ \mathrm{constant}$$