What is the integral of (sin^2(3x)) ?

The integral of (sin^2(3x)) is 1/2 (x-1/6 sin(6x))+C

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integral of (sin^2(3x))

\int (sin^{2} (3 x)) d x

\frac{1}{2} (x - \frac{1}{6} sin (6 x)) + C

Solution steps

\int sin^{2} (3 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (6 x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (6 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d x - \int cos (6 x) d x)

\int 1 d x = x

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

= \frac{1}{2} (x - \frac{1}{6} sin (6 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{6} sin (6 x)) + C

\int 4 x^{2} (4 x^{3} + 5)^{5} d x

ln (3 \frac{3 x}{x + 2})

\int - z^{2} d z

\frac{\partial}{\partial x} (y \cdot e^{\frac{x}{y}})

x \to 0 lim (\frac{sin ( 2 x ) sin ( 3 x )}{sin ( 2 x ) sin ( 3 x )})

What is the integral of (sin^2(3x)) ?
The integral of (sin^2(3x)) is 1/2 (x-1/6 sin(6x))+C