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

The integral of usin^2(3u) is 1/72 (18u^2-6usin(6u)-cos(6u))+C

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integral of usin^2(3u)

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

\frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u)) + C

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

: u (u - \frac{1}{6} sin (6 u)) - \int u - \frac{1}{6} sin (6 u) d u

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

\int u - \frac{1}{6} sin (6 u) d u = \frac{u ^{2}}{2} + \frac{1}{36} cos (6 u)

= \frac{1}{2} (u (u - \frac{1}{6} sin (6 u)) - (\frac{u ^{2}}{2} + \frac{1}{36} cos (6 u)))

Simplify

\frac{1}{2} (u (u - \frac{1}{6} sin (6 u)) - (\frac{u ^{2}}{2} + \frac{1}{36} cos (6 u))) : \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u))

= \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u))

Add a constant to the solution

= \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u)) + C

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What is the integral of usin^2(3u) ?
The integral of usin^2(3u) is 1/72 (18u^2-6usin(6u)-cos(6u))+C