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

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

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integral of (2x-1)y^2sin(3xy)

\int (2 x - 1) y^{2} sin (3 x y) d x

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

Solution steps

\int (2 x - 1) y^{2} sin (3 x y) d x

Take the constant out:

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

= y^{2} \cdot \int (- 1 + 2 x) sin (3 y x) d x

Expand

(- 1 + 2 x) sin (3 y x) : - sin (3 y x) + 2 x sin (3 y x)

= y^{2} \cdot \int - sin (3 y x) + 2 x sin (3 y 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

= y^{2} (- \int sin (3 y x) d x + \int 2 x sin (3 y x) d x)

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

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

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

Simplify

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

Add a constant to the solution

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

\int 15 (tan (x)) (ln (cos (x))) d x

\int \frac{2 x}{x ^{2} - 4} d x

\int - cos (x) + 1 d x

x \to 0 lim (\frac{( x ^{2} + 2 x )}{x})

\frac{\partial}{\partial y} (- 2 x y (- 15 - x - y))

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