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anonymous
 4 years ago
evaluate the indefinite integral. problem below
anonymous
 4 years ago
evaluate the indefinite integral. problem below

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits ( \Theta  \cos ( 1\Theta) d \Theta \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01sin(1theta)/1+C =1 +sin(1theta)+c

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1use integral subs, let u=(1Θ)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops its I did a mistake there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it should be \[\theta ^{2}/2 + \sin(1\theta)+\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if i use subs what do i do after i get du=1d\[\Theta \] ?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1well, i want explain it step by step u = 1Θ > Θ=1u, right ?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1next, u have got u = 1Θ du = dΘ or dΘ =  du, right ?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1next, we substitute of them to the original problem : ∫(Θ−cos(1−Θ))dΘ = ∫[(1u)  cosu] (du) = ∫[(u1) + cosu] du, agree ???

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1ok, just integral all one by one, int u du= ... int 1du= .... int cosu du= .... what u get ????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/2 u^2 u + sin u + c ?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, correct now, the last step u must substitute back that u = 1Θ

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i end up getting 1/2 1/2theta^2 + sin (1theta) is the 1/2 suppose to be there?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1from ur answer : 1/2 u^2 u + sin u + c , just change u=1Θ, gives 1/2(1Θ)^2 (1Θ) + sin(1Θ) + c i think enough be the answer, but if u want simplify it might too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whats the final answer if you simplify?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.11/2(1Θ)^2 (1Θ) + sin(1Θ) + c = 1/2 (12Θ + Θ^2) 1 + Θ + sin(1Θ) + c = 1/2  Θ +1/2*Θ^2 1 + Θ + sin(1Θ) + c = 1/2 +1/2*Θ^2 + sin(1Θ) + c = 1/2 (1Θ^2) + sin(1Θ) + c
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