Integral of tag^5xsec^2xdx dx

1. 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 $du$:

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

      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:

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

   Method #2

   1. Rewrite the integrand:

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

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

      Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ 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:

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

2. Add the constant of integration:

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

The answer is:

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