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\[\int\limits ( \Theta - \cos ( 1-\Theta) d \Theta \]

1-sin(1-theta)/-1+C
=1 +sin(1-theta)+c

use integral subs, let u=(1-Θ)

oops its I did a mistake there

it should be \[\theta ^{2}/2 + \sin(1-\theta)+\]

if i use subs what do i do after i get du=-1d\[\Theta \] ?

well, i want explain it step by step
u = 1-Θ -----> Θ=1-u, right ?

yes

next, u have got
u = 1-Θ
du = -dΘ or
dΘ = - du, right ?

yesss

yes

ok, just integral all one by one,
int u du= ...
int -1du= ....
int cosu du= ....
what u get ????

1/2 u^2 -u + sin u + c ?

yes, correct
now, the last step u must substitute back that u = 1-Θ

i end up getting -1/2 -1/2theta^2 + sin (1-theta) is the -1/2 suppose to be there?

whats the final answer if you simplify?