Question

Integrate Sin(3x)Cos(3x) dx

Answer 1

Answer:

$I = \frac{1}{6}\cdot \sin^{2} 3x + C$

Where $C$ is the integration constant.

Step-by-step explanation:

We use integration by substitution to obtain the integral, where:

$u = \sin 3x$, $du = 3\cdot \cos 3x\,dx$

$I = \int {\sin 3x\cdot \cos 3x} \, dx$

$I = \frac{1}{3} \int {u} \, du$

$I = \frac{1}{6}\cdot u^{2} + C$

$I = \frac{1}{6}\cdot \sin^{2} 3x + C$

Where $C$ is the integration constant.