Question

how do you find the antiderivative of cos(pix)?

Answer 1

i wolframalphad the answer and it's supposed to be sin(pix)/pi but i'm not sure how they got there

Answer 2

You really gotta spend some time and get used to this simple integration trick!
ok ok ok so...

Answer 3

\(\Large\rm (\sin(\pi x))'=\pi\cos(\pi x)\)
When you differentiate, a factor of pi comes out, because of chain rule, ya?

Answer 4

So when you go backwards, what is happening?
You're actually losing a pi, ya?
You're dividing a pi out.

Answer 5

Yes, you can do a u-substitution if you're not comfortable with that shortcut just yet.\[\Large\rm u=\pi x,\qquad du=\pi dx\qquad\to\qquad \frac{1}{\pi}du=dx\]
\[\Large\rm \int\limits\limits \cos(\pi x)dx=\int\limits \cos(u) \left(\frac{1}{\pi}du\right)\]

Answer 6

But try to get comfortable with it! :)
Works the same way with exponentials and stuff:\[\Large\rm \left(e^{2x}\right)'=2e^{2x}\]Therefore,\[\Large\rm \int\limits e^{2x}dx=\frac{1}{2}e^{2x}\]

Answer 7

Thoughts? :o
Confused still?

Answer 8

no i think i got it. thanks for the short cut!! :)

Answer 9

Keep in mind this only works with coefficients on linear x.
Example:\[\Large\rm \frac{d}{dx}\sin(x^2)=2x \cos(x^2)\]Doesn't work out the way we would like with our shortcut,\[\Large\rm \int\limits \cos(x^2)dx\ne\frac{1}{2x}\sin(x^2)\]

Answer 10

ah okay good to know