Integral of cos(3x-2) dx
The solution
Detail solution
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Let u=3x−2.
Then let du=3dx and substitute 3du:
∫9cos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫3cos(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−2)
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Now simplify:
3sin(3x−2)
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Add the constant of integration:
3sin(3x−2)+constant
The answer is:
3sin(3x−2)+constant
The answer (Indefinite)
[src]
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| sin(3*x - 2)
| cos(3*x - 2) dx = C + ------------
| 3
/
3sin(3x−2)
The graph
Use the examples entering the upper and lower limits of integration.