Integral of cos^3(4x) dx

1. Rewrite the integrand:

   $$\cos^{3}{\left(4 x \right)} = \left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)}$$

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sin{\left(4 x \right)}$.

      Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $du$:

      $$\int \left(\frac{1}{4} - \frac{u^{2}}{4}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \frac{1}{4}\, du = \frac{u}{4}$$

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

            $$\int \left(- \frac{u^{2}}{4}\right)\, du = - \frac{\int u^{2}\, du}{4}$$

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

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

            So, the result is: $- \frac{u^{3}}{12}$

         The result is: $- \frac{u^{3}}{12} + \frac{u}{4}$

      Now substitute $u$ back in:

      $$- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         1. Let $u = \sin{\left(4 x \right)}$.

            Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $\frac{du}{4}$:

            $$\int \frac{u^{2}}{16}\, du$$

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

               $$\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}$$

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

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

               So, the result is: $\frac{u^{3}}{12}$

            Now substitute $u$ back in:

            $$\frac{\sin^{3}{\left(4 x \right)}}{12}$$

         So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12}$

      1. Let $u = 4 x$.

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

         $$\int \frac{\cos{\left(u \right)}}{16}\, du$$

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

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

            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:

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

      The result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$

   Method #3

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         1. Let $u = \sin{\left(4 x \right)}$.

            Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $\frac{du}{4}$:

            $$\int \frac{u^{2}}{16}\, du$$

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

               $$\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}$$

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

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

               So, the result is: $\frac{u^{3}}{12}$

            Now substitute $u$ back in:

            $$\frac{\sin^{3}{\left(4 x \right)}}{12}$$

         So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12}$

      1. Let $u = 4 x$.

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

         $$\int \frac{\cos{\left(u \right)}}{16}\, du$$

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

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

            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:

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

      The result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$

3. Now simplify:

   $$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}$$

4. Add the constant of integration:

   $$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}$$