## heydayana evaluate the indefinite integral. problem below one year ago one year ago

1. heydayana

$\int\limits ( \Theta - \cos ( 1-\Theta) d \Theta$

2. artix_17

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

use integral subs, let u=(1-Θ)

4. artix_17

oops its I did a mistake there

5. artix_17

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

6. heydayana

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 ?

8. heydayana

yes

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

10. heydayana

yesss

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

12. heydayana

yes

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

14. heydayana

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

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

16. heydayana

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

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

18. heydayana

whats the final answer if you simplify?