Question

evaluate the indefinite integral. problem below

Answer 1

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

Answer 2

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

Answer 3

use integral subs, let u=(1-Θ)

Answer 4

oops its I did a mistake there

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yes

Answer 9

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

Answer 10

yesss

Answer 11

next, we substitute of them to the original problem :
∫(Θ−cos(1−Θ))dΘ
= ∫[(1-u) - cosu] (-du)
= ∫[(u-1) + cosu] du, agree ???

Answer 12

yes

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

from ur answer :
1/2 u^2 -u + sin u + c , just change u=1-Θ, gives
1/2(1-Θ)^2 -(1-Θ) + sin(1-Θ) + c
i think enough be the answer, but if u want simplify it might too

Answer 18

whats the final answer if you simplify?

Answer 19

1/2(1-Θ)^2 -(1-Θ) + sin(1-Θ) + c
= 1/2 (1-2Θ + Θ^2) -1 + Θ + sin(1-Θ) + c
= 1/2 - Θ +1/2*Θ^2 -1 + Θ + sin(1-Θ) + c
= -1/2 +1/2*Θ^2 + sin(1-Θ) + c
= -1/2 (1-Θ^2) + sin(1-Θ) + c