What is the integral of 9(3x-1)^2cos(3x) ?

The integral of 9(3x-1)^2cos(3x) is 9(3x^2sin(3x)+2xcos(3x)-2xsin(3x)-1/3 sin(3x)-2/3 cos(3x))+C

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integral of 9(3x-1)^2cos(3x)

\int 9 (3 x - 1)^{2} cos (3 x) d x

9 (3 x^{2} sin (3 x) + 2 x cos (3 x) - 2 x sin (3 x) - \frac{1}{3} sin (3 x) - \frac{2}{3} cos (3 x)) + C

Solution steps

\int 9 (3 x - 1)^{2} cos (3 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int (3 x - 1)^{2} cos (3 x) d x

Expand

(3 x - 1)^{2} cos (3 x) : 9 x^{2} cos (3 x) - 6 x cos (3 x) + cos (3 x)

= 9 \cdot \int 9 x^{2} cos (3 x) - 6 x cos (3 x) + cos (3 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int 9 x^{2} cos (3 x) d x - \int 6 x cos (3 x) d x + \int cos (3 x) d x)

\int 9 x^{2} cos (3 x) d x = \frac{1}{3} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

\int 6 x cos (3 x) d x = 2 x sin (3 x) + \frac{2}{3} cos (3 x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

= 9 (\frac{1}{3} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) - (2 x sin (3 x) + \frac{2}{3} cos (3 x)) + \frac{1}{3} sin (3 x))

Expand

\frac{1}{3} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) - (2 x sin (3 x) + \frac{2}{3} cos (3 x)) + \frac{1}{3} sin (3 x) : 3 x^{2} sin (3 x) + 2 x cos (3 x) - 2 x sin (3 x) - \frac{1}{3} sin (3 x) - \frac{2}{3} cos (3 x)

= 9 (3 x^{2} sin (3 x) + 2 x cos (3 x) - 2 x sin (3 x) - \frac{1}{3} sin (3 x) - \frac{2}{3} cos (3 x))

Add a constant to the solution

= 9 (3 x^{2} sin (3 x) + 2 x cos (3 x) - 2 x sin (3 x) - \frac{1}{3} sin (3 x) - \frac{2}{3} cos (3 x)) + C

\int \frac{x}{( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

\int x^{5} \cdot sin (x^{6}) d x

\int 1 cos (nπ x) d x

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

\int \frac{- 9 x ^{2} + 16}{x ^{2}} d x

What is the integral of 9(3x-1)^2cos(3x) ?
The integral of 9(3x-1)^2cos(3x) is 9(3x^2sin(3x)+2xcos(3x)-2xsin(3x)-1/3 sin(3x)-2/3 cos(3x))+C