How do you find the antiderivative of #cos(5x)#?

1 Answer
Oct 28, 2016

Say that:

#y=sin(kx)# whereby k is a constant.

Now, transform this into:

#y=sin(u)# whereby #u=kx#.

If this is the case:

#(dy)/(du)=cos(u)=cos(kx)#

#(du)/(dx)=k#

And this means that:

#(dy)/(du)*(du)/(dx)=(dy)/(dx)=kcos(kx)#

Now, when k=5:

When #y=sin(5x)#, #(dy)/(dx)=5cos(5x)#.

Ok... So what happens when:

#y=1/5sin(5x)#??

Well...

#5y=sin(5x)#

Differentiating both sides of the equation and remembering that #(dy)/(dy)*(dy)/(dx)=(dy)/(dx)# you get...

#5*(dy)/(dx)=5cos(5x)#

Then dividing both sides of this equation by 5 you get...

#(dy)/(dx)=cos(5x)#

This means that:

#intcos(5x)dx=1/5sin(5x)+C#