My calculation:
u=1-cos3t
du/dt=sin3t
du=sin3t dt
t=0, u=1-cos3(0)=1
t=pi/6, u=1-cos3(pi/6)=0.477
integral (1&0.477) u du
(u^2/2)(1&0.477)
then stucked
integrate (pi/6 & 0) (1-cos3t)(sin3t)dt
3 answers
by "inspection" I get
(1/6)(1 - cos 3t)^2 | from 0 to π/6
= (1/6)(1 - 0)^2 - (1/6)(1 - 1)^2
= 1/6 - 0
= 1/6
I was assuming that your limit went from 0 to π/6
(1/6)(1 - cos 3t)^2 | from 0 to π/6
= (1/6)(1 - 0)^2 - (1/6)(1 - 1)^2
= 1/6 - 0
= 1/6
I was assuming that your limit went from 0 to π/6
Still not very understand.
Where u get the 1/6?
Where u get the 1/6?
2 answers