Answer:
[tex]I = \frac{1}{6}\cdot \sin^{2} 3x + C[/tex]
Where [tex]C[/tex] is the integration constant.
Step-by-step explanation:
We use integration by substitution to obtain the integral, where:
[tex]u = \sin 3x[/tex], [tex]du = 3\cdot \cos 3x\,dx[/tex]
[tex]I = \int {\sin 3x\cdot \cos 3x} \, dx[/tex]
[tex]I = \frac{1}{3} \int {u} \, du[/tex]
[tex]I = \frac{1}{6}\cdot u^{2} + C[/tex]
[tex]I = \frac{1}{6}\cdot \sin^{2} 3x + C[/tex]
Where [tex]C[/tex] is the integration constant.
