Integral of tan(5x)^5 dx

1. Rewrite the integrand:

   $$\tan^{5}{\left(5 x \right)} = \left(\sec^{2}{\left(5 x \right)} - 1\right)^{2} \tan{\left(5 x \right)}$$

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sec^{2}{\left(5 x \right)}$.

      Then let $du = 10 \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)} dx$ and substitute $\frac{du}{10}$:

      $$\int \frac{u^{2} - 2 u + 1}{10 u}\, du$$

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

         $$\int \frac{u^{2} - 2 u + 1}{u}\, du = \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{10}$$

         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 $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

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

               $$\int \left(-2\right)\, du = - 2 u$$

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

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

         So, the result is: $\frac{u^{2}}{20} - \frac{u}{5} + \frac{\log{\left(u \right)}}{10}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(\sec^{2}{\left(5 x \right)} \right)}}{10} + \frac{\sec^{4}{\left(5 x \right)}}{20} - \frac{\sec^{2}{\left(5 x \right)}}{5}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(\sec^{2}{\left(5 x \right)} - 1\right)^{2} \tan{\left(5 x \right)} = \tan{\left(5 x \right)} \sec^{4}{\left(5 x \right)} - 2 \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)} + \tan{\left(5 x \right)}$$

   2. Integrate term-by-term:

      1. Let $u = \sec{\left(5 x \right)}$.

         Then let $du = 5 \tan{\left(5 x \right)} \sec{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$:

         $$\int \frac{u^{3}}{5}\, du$$

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

            $$\int u^{3}\, du = \frac{\int u^{3}\, du}{5}$$

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

               $$\int u^{3}\, du = \frac{u^{4}}{4}$$

            So, the result is: $\frac{u^{4}}{20}$

         Now substitute $u$ back in:

         $$\frac{\sec^{4}{\left(5 x \right)}}{20}$$

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

         $$\int \left(- 2 \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)}\right)\, dx = - 2 \int \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)}\, dx$$

         1. Let $u = \sec{\left(5 x \right)}$.

            Then let $du = 5 \tan{\left(5 x \right)} \sec{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$:

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

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

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

               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: $\frac{u^{2}}{10}$

            Now substitute $u$ back in:

            $$\frac{\sec^{2}{\left(5 x \right)}}{10}$$

         So, the result is: $- \frac{\sec^{2}{\left(5 x \right)}}{5}$

      1. Rewrite the integrand:

         $$\tan{\left(5 x \right)} = \frac{\sin{\left(5 x \right)}}{\cos{\left(5 x \right)}}$$

      2. Let $u = \cos{\left(5 x \right)}$.

         Then let $du = - 5 \sin{\left(5 x \right)} dx$ and substitute $- \frac{du}{5}$:

         $$\int \left(- \frac{1}{5 u}\right)\, du$$

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

            $$\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{5}$$

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

            So, the result is: $- \frac{\log{\left(u \right)}}{5}$

         Now substitute $u$ back in:

         $$- \frac{\log{\left(\cos{\left(5 x \right)} \right)}}{5}$$

      The result is: $- \frac{\log{\left(\cos{\left(5 x \right)} \right)}}{5} + \frac{\sec^{4}{\left(5 x \right)}}{20} - \frac{\sec^{2}{\left(5 x \right)}}{5}$

   Method #3

   1. Rewrite the integrand:

      $$\left(\sec^{2}{\left(5 x \right)} - 1\right)^{2} \tan{\left(5 x \right)} = \tan{\left(5 x \right)} \sec^{4}{\left(5 x \right)} - 2 \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)} + \tan{\left(5 x \right)}$$

   2. Integrate term-by-term:

      1. Let $u = \sec{\left(5 x \right)}$.

         Then let $du = 5 \tan{\left(5 x \right)} \sec{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$:

         $$\int \frac{u^{3}}{5}\, du$$

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

            $$\int u^{3}\, du = \frac{\int u^{3}\, du}{5}$$

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

               $$\int u^{3}\, du = \frac{u^{4}}{4}$$

            So, the result is: $\frac{u^{4}}{20}$

         Now substitute $u$ back in:

         $$\frac{\sec^{4}{\left(5 x \right)}}{20}$$

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

         $$\int \left(- 2 \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)}\right)\, dx = - 2 \int \tan{\left(5 x \right)} \sec^{2}{\left(5 x \right)}\, dx$$

         1. Let $u = \sec{\left(5 x \right)}$.

            Then let $du = 5 \tan{\left(5 x \right)} \sec{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$:

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

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

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

               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: $\frac{u^{2}}{10}$

            Now substitute $u$ back in:

            $$\frac{\sec^{2}{\left(5 x \right)}}{10}$$

         So, the result is: $- \frac{\sec^{2}{\left(5 x \right)}}{5}$

      1. Rewrite the integrand:

         $$\tan{\left(5 x \right)} = \frac{\sin{\left(5 x \right)}}{\cos{\left(5 x \right)}}$$

      2. Let $u = \cos{\left(5 x \right)}$.

         Then let $du = - 5 \sin{\left(5 x \right)} dx$ and substitute $- \frac{du}{5}$:

         $$\int \left(- \frac{1}{5 u}\right)\, du$$

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

            $$\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{5}$$

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

            So, the result is: $- \frac{\log{\left(u \right)}}{5}$

         Now substitute $u$ back in:

         $$- \frac{\log{\left(\cos{\left(5 x \right)} \right)}}{5}$$

      The result is: $- \frac{\log{\left(\cos{\left(5 x \right)} \right)}}{5} + \frac{\sec^{4}{\left(5 x \right)}}{20} - \frac{\sec^{2}{\left(5 x \right)}}{5}$

3. Add the constant of integration:

   $$\frac{\log{\left(\sec^{2}{\left(5 x \right)} \right)}}{10} + \frac{\sec^{4}{\left(5 x \right)}}{20} - \frac{\sec^{2}{\left(5 x \right)}}{5}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(\sec^{2}{\left(5 x \right)} \right)}}{10} + \frac{\sec^{4}{\left(5 x \right)}}{20} - \frac{\sec^{2}{\left(5 x \right)}}{5}+ \mathrm{constant}$$