anonymous
  • anonymous
evaluate the indefinite integral. problem below
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits ( \Theta - \cos ( 1-\Theta) d \Theta \]
anonymous
  • anonymous
1-sin(1-theta)/-1+C =1 +sin(1-theta)+c
RadEn
  • RadEn
use integral subs, let u=(1-Θ)

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anonymous
  • anonymous
oops its I did a mistake there
anonymous
  • anonymous
it should be \[\theta ^{2}/2 + \sin(1-\theta)+\]
anonymous
  • anonymous
if i use subs what do i do after i get du=-1d\[\Theta \] ?
RadEn
  • RadEn
well, i want explain it step by step u = 1-Θ -----> Θ=1-u, right ?
anonymous
  • anonymous
yes
RadEn
  • RadEn
next, u have got u = 1-Θ du = -dΘ or dΘ = - du, right ?
anonymous
  • anonymous
yesss
RadEn
  • RadEn
next, we substitute of them to the original problem : ∫(Θ−cos(1−Θ))dΘ = ∫[(1-u) - cosu] (-du) = ∫[(u-1) + cosu] du, agree ???
anonymous
  • anonymous
yes
RadEn
  • RadEn
ok, just integral all one by one, int u du= ... int -1du= .... int cosu du= .... what u get ????
anonymous
  • anonymous
1/2 u^2 -u + sin u + c ?
RadEn
  • RadEn
yes, correct now, the last step u must substitute back that u = 1-Θ
anonymous
  • anonymous
i end up getting -1/2 -1/2theta^2 + sin (1-theta) is the -1/2 suppose to be there?
RadEn
  • RadEn
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
anonymous
  • anonymous
whats the final answer if you simplify?
RadEn
  • RadEn
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

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