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heydayana
evaluate the indefinite integral. problem below
\[\int\limits ( \Theta - \cos ( 1-\Theta) d \Theta \]
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 ?
next, u have got u = 1-Θ du = -dΘ or dΘ = - du, right ?
next, we substitute of them to the original problem : ∫(Θ−cos(1−Θ))dΘ = ∫[(1-u) - cosu] (-du) = ∫[(u-1) + cosu] du, agree ???
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?
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
whats the final answer if you simplify?
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