What is the integral of xcos^2(9x) ?

The integral of xcos^2(9x) is 1/648 (162x^2+18xsin(18x)+cos(18x))+C

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integral of xcos^2(9x)

\int x cos^{2} (9 x) d x

\frac{1}{648} (162 x^{2} + 18 x sin (18 x) + cos (18 x)) + C

Solution steps

\int x cos^{2} (9 x) d x

Rewrite using trig identities

= \int \frac{1}{2} x (1 + cos (18 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 (18 x)) d x

Apply Integration By Parts

: x (x + \frac{1}{18} sin (18 x)) - \int x + \frac{1}{18} sin (18 x) d x

= \frac{1}{2} (x (x + \frac{1}{18} sin (18 x)) - \int x + \frac{1}{18} sin (18 x) d x)

\int x + \frac{1}{18} sin (18 x) d x = \frac{x ^{2}}{2} - \frac{1}{324} cos (18 x)

= \frac{1}{2} (x (x + \frac{1}{18} sin (18 x)) - (\frac{x ^{2}}{2} - \frac{1}{324} cos (18 x)))

Simplify

\frac{1}{2} (x (x + \frac{1}{18} sin (18 x)) - (\frac{x ^{2}}{2} - \frac{1}{324} cos (18 x))) : \frac{1}{648} (162 x^{2} + 18 x sin (18 x) + cos (18 x))

= \frac{1}{648} (162 x^{2} + 18 x sin (18 x) + cos (18 x))

Add a constant to the solution

= \frac{1}{648} (162 x^{2} + 18 x sin (18 x) + cos (18 x)) + C

\frac{d y}{d x} = 4 x e^{y}

\frac{d}{d x} ((x^{3} + 1) (3 - 2 x^{3}))

(3, 2), (5, 6)

y = 4 x, y = \frac{7}{5} x, y = 32 - x^{2}

\frac{d}{d x} (\frac{x - 5 x x}{x})

What is the integral of xcos^2(9x) ?
The integral of xcos^2(9x) is 1/648 (162x^2+18xsin(18x)+cos(18x))+C