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