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_{- 1}^{2} x d x

cos (x + \frac{π}{2})

\int_{- 1}^{1} \frac{1}{2} (6 - 6 x^{2})^{2} d x

\int t (t + 1)^{6} d t

f (x) = x^{2} - 1, at x = 2

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