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

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

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integral of sin^2(3t)

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

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

Solution steps

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

Rewrite using trig identities

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

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 t) d t

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 t - \int cos (6 t) d t)

\int 1 d t = t

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

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

Add a constant to the solution

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

\int 5 sin (ln (x)) d x

f (x) = \frac{x}{x ^{2} + 36}

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

2 ln (x + 1)

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