What is the integral of (2x-1)sin(3x) ?

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

integral of (2x-1)sin(3x)

\int (2 x - 1) sin (3 x) d x

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

Solution steps

\int (2 x - 1) sin (3 x) d x

Apply Integration By Parts

= - \frac{1}{3} cos (3 x) (2 x - 1) - \int - \frac{2}{3} cos (3 x) d x

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

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

Simplify

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

Add a constant to the solution

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

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

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 4 e^{- x}

d er i v a t i v e cos^{2} (2 x)

\int \frac{x ^{5}}{x ^{3} - 8} d x

d er i v a t i v e f (x) = - 4.9 x^{2} + 35 x + 4

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