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

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

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integral of xcos^2(3x)

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

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

Solution steps

\int x cos^{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 (1 - t)^{2} d t

\frac{d}{d x} ((x^{3} - x) (2 x - x^{2}))

2 y + \frac{d y}{d x} = y^{2} + 1

h (x) = \frac{3}{x ^{2}} + 10 x^{3} - 2 sin (3 x)

\int \frac{2 + 3 x}{x - 3} d x

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