What is the integral of (12ycos(4x)-3,y) ?

The integral of (12ycos(4x)-3,y) is -(,3y)x+3ysin(4x)+C

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integral of (12ycos(4x)-3,y)

\int (12 y cos (4 x) - 3, y) d x

- (, 3 y) x + 3 y sin (4 x) + C

Solution steps

\int 12 y cos (4 x) - 3, y 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, 3 y d x + \int 12 cos (4 x) y d x

\int, 3 y d x = (, 3 y) x

\int 12 cos (4 x) y d x = 3 y sin (4 x)

= - (, 3 y) x + 3 y sin (4 x)

Add a constant to the solution

= - (, 3 y) x + 3 y sin (4 x) + C

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What is the integral of (12ycos(4x)-3,y) ?
The integral of (12ycos(4x)-3,y) is -(,3y)x+3ysin(4x)+C