Integral of sec^4(x) dx

1. Rewrite the integrand:

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

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 $du$:

      $$\int \left(u^{2} + 1\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

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

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

         The result is: $\frac{u^{3}}{3} + u$

      Now substitute $u$ back in:

      $$\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

   Method #2

   1. Rewrite the integrand:

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

   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 $du$:

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

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

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         Now substitute $u$ back in:

         $$\frac{\tan^{3}{\left(x \right)}}{3}$$

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

      The result is: $\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$

   Method #3

   1. Rewrite the integrand:

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

   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 $du$:

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

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

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         Now substitute $u$ back in:

         $$\frac{\tan^{3}{\left(x \right)}}{3}$$

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

      The result is: $\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$

3. Add the constant of integration:

   $$\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+ \mathrm{constant}$$