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

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

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integral of (2x+1)y^2sin(2xy)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x y sin (x) + y e^{x y})

f (x) = \frac{5 x - 7}{6 x - 9}

\int e^{x} 29 + e^{x} d x

\int x e^{t x} d x

\frac{d y}{d t} = 7 t e^{- y}, y (0) = 3

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