What is the integral of 1/2 sin(7y)-1/2 sin(3y) ?

The integral of 1/2 sin(7y)-1/2 sin(3y) is -1/14 cos(7y)+1/6 cos(3y)+C

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integral of 1/2 sin(7y)-1/2 sin(3y)

\int \frac{1}{2} sin (7 y) - \frac{1}{2} sin (3 y) d y

- \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y) + C

Solution steps

\int \frac{1}{2} sin (7 y) - \frac{1}{2} sin (3 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 \frac{1}{2} sin (7 y) d y - \int \frac{1}{2} sin (3 y) d y

\int \frac{1}{2} sin (7 y) d y = - \frac{1}{14} cos (7 y)

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

= - \frac{1}{14} cos (7 y) - (- \frac{1}{6} cos (3 y))

Simplify

= - \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y)

Add a constant to the solution

= - \frac{1}{14} cos (7 y) + \frac{1}{6} cos (3 y) + C

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What is the integral of 1/2 sin(7y)-1/2 sin(3y) ?
The integral of 1/2 sin(7y)-1/2 sin(3y) is -1/14 cos(7y)+1/6 cos(3y)+C