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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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