Integral of cos(3x-5) dx
The solution
Detail solution
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Let u=3x−5.
Then let du=3dx and substitute 3du:
∫3cos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=3∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 3sin(u)
Now substitute u back in:
3sin(3x−5)
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Now simplify:
3sin(3x−5)
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Add the constant of integration:
3sin(3x−5)+constant
The answer is:
3sin(3x−5)+constant
The answer (Indefinite)
[src]
/
| sin(3*x - 5)
| cos(3*x - 5) dx = C + ------------
| 3
/
∫cos(3x−5)dx=C+3sin(3x−5)
The graph
1 sin(5) 1 sin(5)
<- - - ------, - - ------>
3 3 3 3
⟨−31−3sin(5),31−3sin(5)⟩
=
1 sin(5) 1 sin(5)
<- - - ------, - - ------>
3 3 3 3
⟨−31−3sin(5),31−3sin(5)⟩
AccumBounds(-1/3 - sin(5)/3, 1/3 - sin(5)/3)
Use the examples entering the upper and lower limits of integration.