Integral of cos(4x-3)dx dx
The solution
Detail solution
-
Let u=4x−3.
Then let du=4dx and substitute 4du:
∫4cos(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=4∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 4sin(u)
Now substitute u back in:
4sin(4x−3)
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Now simplify:
4sin(4x−3)
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Add the constant of integration:
4sin(4x−3)+constant
The answer is:
4sin(4x−3)+constant
The answer (Indefinite)
[src]
/
| sin(4*x - 3)
| cos(4*x - 3) dx = C + ------------
| 4
/
∫cos(4x−3)dx=C+4sin(4x−3)
The graph
sin(1) sin(3)
------ + ------
4 4
4sin(3)+4sin(1)
=
sin(1) sin(3)
------ + ------
4 4
4sin(3)+4sin(1)
Use the examples entering the upper and lower limits of integration.