What is the integral of ((2x-1)y^2sin(4xy)) ?

The integral of ((2x-1)y^2sin(4xy)) is y^2(1/(4y)cos(4yx)+2(-1/(4y)xcos(4yx)+1/(16y^2)sin(4yx)))+C

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integral of ((2x-1)y^2sin(4xy))

\int ((2 x - 1) y^{2} sin (4 x y)) d x

y^{2} (\frac{1}{4 y} cos (4 y x) + 2 (- \frac{1}{4 y} x cos (4 y x) + \frac{1}{16 y ^{2}} sin (4 y x))) + C

Solution steps

\int (2 x - 1) y^{2} sin (4 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 (4 y x) d x

Expand

(- 1 + 2 x) sin (4 y x) : - sin (4 y x) + 2 x sin (4 y x)

= y^{2} \cdot \int - sin (4 y x) + 2 x sin (4 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 (4 y x) d x + \int 2 x sin (4 y x) d x)

\int sin (4 y x) d x = - \frac{1}{4 y} cos (4 y x)

\int 2 x sin (4 y x) d x = 2 (- \frac{1}{4 y} x cos (4 y x) + \frac{1}{16 y ^{2}} sin (4 y x))

= y^{2} (- (- \frac{1}{4 y} cos (4 y x)) + 2 (- \frac{1}{4 y} x cos (4 y x) + \frac{1}{16 y ^{2}} sin (4 y x)))

Simplify

= y^{2} (\frac{1}{4 y} cos (4 y x) + 2 (- \frac{1}{4 y} x cos (4 y x) + \frac{1}{16 y ^{2}} sin (4 y x)))

Add a constant to the solution

= y^{2} (\frac{1}{4 y} cos (4 y x) + 2 (- \frac{1}{4 y} x cos (4 y x) + \frac{1}{16 y ^{2}} sin (4 y x))) + C

\int 17 x e^{19 x} d x

x \to 0 + lim ((tan (5 x))^{x})

\frac{d}{d x} (- \frac{e ^{x}}{x})

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

\frac{d}{d x} (\frac{5}{x ^{9}})

What is the integral of ((2x-1)y^2sin(4xy)) ?
The integral of ((2x-1)y^2sin(4xy)) is y^2(1/(4y)cos(4yx)+2(-1/(4y)xcos(4yx)+1/(16y^2)sin(4yx)))+C