What is the antiderivative of #sin(8(pi)x)dx#?

Question

2153 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-06T14:32:30

#-1/(8pi)cos(8pix) +C#

From our experience with finding derivative, we would expect the anti derivative to involve #cos(8pix)#

But the derivative of #cos(8pix)# is

#d/dx(cos(8pix)) = -sin(8pix) * d/dx(8pix)#

# = -sin(8pix) * (8pi)#

# = -8pisin(8pix)#.

This is a constant multiple of the function we started with.

We can "correct" this by starting with

#-1/(8pi)cos(8pix)#

Now the derivative is:

#d/dx(-1/(8pi)cos(8pix)) = -1/(8pi) d/dx(cos(8pix)) #

# = -1/(8pi) [-sin(8pix) * (8pi)]#

# = sin(8pix)# And don't forget the constant!