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

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

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

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

\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) sin (4 x y) d x

Expand

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

= \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

= - \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))

= - (- \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

= \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

= \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

\frac{2}{t ^{4}}

n = 1 \sum \infty \frac{( 1 )}{2 n + 7}

n = 2 \sum \infty 1 - \frac{1}{n ^{2}}

y = x^{4} ln (x)

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

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