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

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

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integral of xsin^2(3x)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 3 t d t

1 - x x

\frac{\partial}{\partial z} (y cos (z))

3 x + 3 x

\frac{y ^{2}}{2} = x, x = 2 y

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