Integral of 3cos3x dx

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

   $$\int 3 \cos{\left(3 x \right)}\, dx = 3 \int \cos{\left(3 x \right)}\, dx$$

   1. Let $u = 3 x$.

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

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

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

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

         1. The integral of cosine is sine:

            $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

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

      Now substitute $u$ back in:

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

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

2. Add the constant of integration:

   $$\sin{\left(3 x \right)}+ \mathrm{constant}$$

The answer is: