Question

how can i integrate cos2t without using substituion

Answer 1

Realize that if you differntiate it you have to do the chain rule giving -2sin(2t)
So that means if you integrate it you need a factor of (1/2) to take care of the 2 outside.
So the integral of cos(2t)=(1/2)sin(2t)
Because:
\[\frac{d}{dt}(\frac{1}{2})\sin(2t)=\cos(2t)\]

Answer 2

why is it -2sin(2t), why a negative

Answer 3

Because the derivative of cos(x)=-sin(x)

Answer 4

The negative isn't too important
You can use hyperbolic trig functions (they don't have the negative)
\[\frac{d}{dt}\sinh(3t)=3\cosh(3t) \rightarrow \int\limits \frac{d}{dt}\sinh(3t)=3\int\limits \cosh(3t) \rightarrow \frac{1}{3}\sinh(3t)=\int\limits \cosh(3t)\]
You just need the fraction to compensate for the 3t inside the function ("undoing" the chain rule, if you will).