Integral of sec^2xtanx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

         $$\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$$

         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:

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

   Method #2

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

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

      $$\int u\, du$$

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

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

      Now substitute $u$ back in:

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

   Method #3

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

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

      $$\int u\, du$$

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

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

      Now substitute $u$ back in:

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

2. Add the constant of integration:

   $$\frac{\sec^{2}{\left(x \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{\sec^{2}{\left(x \right)}}{2}+ \mathrm{constant}$$