Integral of sin^3(6x) dx

1. Rewrite the integrand:

   $$\sin^{3}{\left(6 x \right)} = \left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)}$$

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \cos{\left(6 x \right)}$.

      Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $du$:

      $$\int \left(\frac{u^{2}}{6} - \frac{1}{6}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \frac{u^{2}}{6}\, du = \frac{\int u^{2}\, du}{6}$$

            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: $\frac{u^{3}}{18}$

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

            $$\int \left(- \frac{1}{6}\right)\, du = - \frac{u}{6}$$

         The result is: $\frac{u^{3}}{18} - \frac{u}{6}$

      Now substitute $u$ back in:

      $$\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)} = - \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)} + \sin{\left(6 x \right)}$$

   2. Integrate term-by-term:

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

         $$\int \left(- \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\right)\, dx = - \int \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\, dx$$

         1. Let $u = \cos{\left(6 x \right)}$.

            Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $- \frac{du}{6}$:

            $$\int \frac{u^{2}}{36}\, du$$

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

               $$\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$$

               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: $- \frac{u^{3}}{18}$

            Now substitute $u$ back in:

            $$- \frac{\cos^{3}{\left(6 x \right)}}{18}$$

         So, the result is: $\frac{\cos^{3}{\left(6 x \right)}}{18}$

      1. Let $u = 6 x$.

         Then let $du = 6 dx$ and substitute $\frac{du}{6}$:

         $$\int \frac{\sin{\left(u \right)}}{36}\, du$$

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

            $$\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$$

            1. The integral of sine is negative cosine:

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

            So, the result is: $- \frac{\cos{\left(u \right)}}{6}$

         Now substitute $u$ back in:

         $$- \frac{\cos{\left(6 x \right)}}{6}$$

      The result is: $\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$

   Method #3

   1. Rewrite the integrand:

      $$\left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)} = - \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)} + \sin{\left(6 x \right)}$$

   2. Integrate term-by-term:

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

         $$\int \left(- \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\right)\, dx = - \int \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\, dx$$

         1. Let $u = \cos{\left(6 x \right)}$.

            Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $- \frac{du}{6}$:

            $$\int \frac{u^{2}}{36}\, du$$

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

               $$\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$$

               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: $- \frac{u^{3}}{18}$

            Now substitute $u$ back in:

            $$- \frac{\cos^{3}{\left(6 x \right)}}{18}$$

         So, the result is: $\frac{\cos^{3}{\left(6 x \right)}}{18}$

      1. Let $u = 6 x$.

         Then let $du = 6 dx$ and substitute $\frac{du}{6}$:

         $$\int \frac{\sin{\left(u \right)}}{36}\, du$$

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

            $$\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$$

            1. The integral of sine is negative cosine:

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

            So, the result is: $- \frac{\cos{\left(u \right)}}{6}$

         Now substitute $u$ back in:

         $$- \frac{\cos{\left(6 x \right)}}{6}$$

      The result is: $\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$

3. Now simplify:

   $$\frac{\left(\cos^{2}{\left(6 x \right)} - 3\right) \cos{\left(6 x \right)}}{18}$$

4. Add the constant of integration:

   $$\frac{\left(\cos^{2}{\left(6 x \right)} - 3\right) \cos{\left(6 x \right)}}{18}+ \mathrm{constant}$$

The answer is: