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

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

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integral of sin^2(3u)

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

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

Solution steps

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

Rewrite using trig identities

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

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 u) d u

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

\int 1 d u = u

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

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

Add a constant to the solution

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

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