ò(2+2cosx)1/2 dx
By Anonymous on Thursday, March 8, 2001
- 06:33 pm :
How do you integrate ò(2+2cosx)1/2 dx?
Thanks
By Dan Goodman (Dfmg2) on Thursday,
March 8, 2001 - 06:47 pm :
You have to notice that cos(2x)=cos2(x)-sin2(x) = 2cos2(x)-1.
So 1+cos(2x)=2cos2(x).
So your integral is just ò(2(1+cos(x))1/2dx=ò(4cos2(x/2))1/2dx = 2òcos(x/2)dx=4sin(x/2)+C where C is a constant.
The things to remember are that cos(2x)=cos2(x)-sin2(x),
sin(2x)=2cos(x)sin(x)
You should also try to remember that 1+cos(2x) can be written as a multiple
of cos2(x) and 1-cos(2x) is a multiple of sin2(x), and then just
quickly work out the exact form when you need it.
By Anonymous on Thursday, March 8, 2001
- 07:19 pm :
Thanks for those tips Dan! ;)