Integral of sin(3*x)*cos(3*x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sin{\left(3 x \right)}$.

      Then let $du = 3 \cos{\left(3 x \right)} dx$ and substitute $\frac{du}{3}$:

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

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

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

         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: $\frac{u^{2}}{6}$

      Now substitute $u$ back in:

      $$\frac{\sin^{2}{\left(3 x \right)}}{6}$$

   Method #2

   1. Let $u = 3 x$.

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

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

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

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

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

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

            $$\int \left(- u\right)\, du$$

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

               $$\int u\, du = - \int u\, du$$

               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: $- \frac{u^{2}}{2}$

            Now substitute $u$ back in:

            $$- \frac{\cos^{2}{\left(u \right)}}{2}$$

         So, the result is: $- \frac{\cos^{2}{\left(u \right)}}{6}$

      Now substitute $u$ back in:

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

   Method #3

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

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

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

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

         $$\int u\, du = - \frac{\int u\, du}{3}$$

         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: $- \frac{u^{2}}{6}$

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is:

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