Integral of tg^4x/cos^4x dx

1. Rewrite the integrand:

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

2. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$:

      $$\int \left(u^{6} + u^{4}\right)\, du$$

      1. Integrate term-by-term:

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

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

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

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         The result is: $\frac{u^{7}}{7} + \frac{u^{5}}{5}$

      Now substitute $u$ back in:

      $$\frac{\tan^{7}{\left(x \right)}}{7} + \frac{\tan^{5}{\left(x \right)}}{5}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(\tan^{2}{\left(x \right)} + 1\right) \tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} = \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)} + \tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)}$$

   2. Integrate term-by-term:

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

         Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$:

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

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

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

         Now substitute $u$ back in:

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

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

         Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$:

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

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

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         Now substitute $u$ back in:

         $$\frac{\tan^{5}{\left(x \right)}}{5}$$

      The result is: $\frac{\tan^{7}{\left(x \right)}}{7} + \frac{\tan^{5}{\left(x \right)}}{5}$

   Method #3

   1. Rewrite the integrand:

      $$\left(\tan^{2}{\left(x \right)} + 1\right) \tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} = \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)} + \tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)}$$

   2. Integrate term-by-term:

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

         Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$:

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

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

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

         Now substitute $u$ back in:

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

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

         Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$:

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

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

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         Now substitute $u$ back in:

         $$\frac{\tan^{5}{\left(x \right)}}{5}$$

      The result is: $\frac{\tan^{7}{\left(x \right)}}{7} + \frac{\tan^{5}{\left(x \right)}}{5}$

3. Add the constant of integration:

   $$\frac{\tan^{7}{\left(x \right)}}{7} + \frac{\tan^{5}{\left(x \right)}}{5}+ \mathrm{constant}$$

The answer is:

$$\frac{\tan^{7}{\left(x \right)}}{7} + \frac{\tan^{5}{\left(x \right)}}{5}+ \mathrm{constant}$$