Integral of 2sec(3x)tan(3x) dx

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

   $$\int 2 \tan{\left(3 x \right)} \sec{\left(3 x \right)}\, dx = 2 \int \tan{\left(3 x \right)} \sec{\left(3 x \right)}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = \sec{\left(3 x \right)}$.

         Then let $du = 3 \tan{\left(3 x \right)} \sec{\left(3 x \right)} dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{1}{9}\, du$$

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

            $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

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

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

            So, the result is: $\frac{u}{3}$

         Now substitute $u$ back in:

         $$\frac{\sec{\left(3 x \right)}}{3}$$

      Method #2

      1. Let $u = 3 x$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{\tan{\left(u \right)} \sec{\left(u \right)}}{9}\, du$$

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

            $$\int \frac{\tan{\left(u \right)} \sec{\left(u \right)}}{3}\, du = \frac{\int \tan{\left(u \right)} \sec{\left(u \right)}\, du}{3}$$

            1. The integral of secant times tangent is secant:

               $$\int \tan{\left(u \right)} \sec{\left(u \right)}\, du = \sec{\left(u \right)}$$

            So, the result is: $\frac{\sec{\left(u \right)}}{3}$

         Now substitute $u$ back in:

         $$\frac{\sec{\left(3 x \right)}}{3}$$

   So, the result is: $\frac{2 \sec{\left(3 x \right)}}{3}$

2. Add the constant of integration:

   $$\frac{2 \sec{\left(3 x \right)}}{3}+ \mathrm{constant}$$

The answer is: