What is the integral of sin(2y)(3-cos(2y))^2 ?

The integral of sin(2y)(3-cos(2y))^2 is -9/2 cos(2y)+3/4 cos(4y)-1/6 cos^3(2y)+C

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integral of sin(2y)(3-cos(2y))^2

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

- \frac{9}{2} cos (2 y) + \frac{3}{4} cos (4 y) - \frac{1}{6} cos^{3} (2 y) + C

Solution steps

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

Expand

sin (2 y) (3 - cos (2 y))^{2} : 9 sin (2 y) - 6 sin (2 y) cos (2 y) + cos^{2} (2 y) sin (2 y)

= \int 9 sin (2 y) - 6 sin (2 y) cos (2 y) + cos^{2} (2 y) sin (2 y) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 sin (2 y) d y - \int 6 sin (2 y) cos (2 y) d y + \int cos^{2} (2 y) sin (2 y) d y

\int 9 sin (2 y) d y = - \frac{9}{2} cos (2 y)

\int 6 sin (2 y) cos (2 y) d y = - \frac{3}{4} cos (4 y)

\int cos^{2} (2 y) sin (2 y) d y = - \frac{1}{6} cos^{3} (2 y)

= - \frac{9}{2} cos (2 y) - (- \frac{3}{4} cos (4 y)) - \frac{1}{6} cos^{3} (2 y)

Simplify

= - \frac{9}{2} cos (2 y) + \frac{3}{4} cos (4 y) - \frac{1}{6} cos^{3} (2 y)

Add a constant to the solution

= - \frac{9}{2} cos (2 y) + \frac{3}{4} cos (4 y) - \frac{1}{6} cos^{3} (2 y) + C

\frac{x - 1}{x + 1}

\int_{0}^{1} 6 x (1 - x) d x

f (x) = - cos (2 x)

\frac{\partial}{\partial x} (\frac{z - x}{z + y})

\frac{d}{d x} (\frac{16 x}{( 4 - x ^{2} ) ^{2}})

What is the integral of sin(2y)(3-cos(2y))^2 ?
The integral of sin(2y)(3-cos(2y))^2 is -9/2 cos(2y)+3/4 cos(4y)-1/6 cos^3(2y)+C