Integral of cos^3(3t)sin(3t)dt dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \cos{\left(3 t \right)}$.

      Then let $du = - 3 \sin{\left(3 t \right)} dt$ and substitute $- \frac{du}{3}$:

      $$\int \frac{u^{3}}{9}\, du$$

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

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

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

            $$\int u^{3}\, du = \frac{u^{4}}{4}$$

         So, the result is: $- \frac{u^{4}}{12}$

      Now substitute $u$ back in:

      $$- \frac{\cos^{4}{\left(3 t \right)}}{12}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Let $u = \sin^{2}{\left(3 t \right)}$.

      Then let $du = 6 \sin{\left(3 t \right)} \cos{\left(3 t \right)} dt$ and substitute $du$:

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

      1. Integrate term-by-term:

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

            $$\int \frac{1}{6}\, du = \frac{u}{6}$$

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

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

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

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

            So, the result is: $- \frac{u^{2}}{12}$

         The result is: $- \frac{u^{2}}{12} + \frac{u}{6}$

      Now substitute $u$ back in:

      $$- \frac{\sin^{4}{\left(3 t \right)}}{12} + \frac{\sin^{2}{\left(3 t \right)}}{6}$$

2. Add the constant of integration:

   $$- \frac{\cos^{4}{\left(3 t \right)}}{12}+ \mathrm{constant}$$

The answer is: