Integral of (cos²(2x)-sin²(2x)) dx

1. Integrate term-by-term:

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

      $$\int \left(- \sin^{2}{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{2}{\left(2 x \right)} = \frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}$$

      2. Integrate term-by-term:

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

            $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

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

            $$\int \left(- \frac{\cos{\left(4 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{2}$$

            1. Let $u = 4 x$.

               Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

               $$\int \frac{\cos{\left(u \right)}}{4}\, du$$

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

                  $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$$

                  1. The integral of cosine is sine:

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

                  So, the result is: $\frac{\sin{\left(u \right)}}{4}$

               Now substitute $u$ back in:

               $$\frac{\sin{\left(4 x \right)}}{4}$$

            So, the result is: $- \frac{\sin{\left(4 x \right)}}{8}$

         The result is: $\frac{x}{2} - \frac{\sin{\left(4 x \right)}}{8}$

      So, the result is: $- \frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$

   1. Rewrite the integrand:

      $$\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$$

   2. Integrate term-by-term:

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

         $$\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$$

         1. Let $u = 4 x$.

            Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

            $$\int \frac{\cos{\left(u \right)}}{4}\, du$$

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

               $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$$

               1. The integral of cosine is sine:

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

               So, the result is: $\frac{\sin{\left(u \right)}}{4}$

            Now substitute $u$ back in:

            $$\frac{\sin{\left(4 x \right)}}{4}$$

         So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$

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

         $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$

   The result is: $\frac{\sin{\left(4 x \right)}}{4}$

2. Add the constant of integration:

   $$\frac{\sin{\left(4 x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{\sin{\left(4 x \right)}}{4}+ \mathrm{constant}$$