Integral of 2sin4x dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫2sin(4x)dx=2∫sin(4x)dx
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Let u=4x.
Then let du=4dx and substitute 4du:
∫16sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫4sin(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)
So, the result is: −2cos(4x)
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Add the constant of integration:
−2cos(4x)+constant
The answer is:
−2cos(4x)+constant
The answer (Indefinite)
[src]
/
| cos(4*x)
| 2*sin(4*x) dx = C - --------
| 2
/
−2cos(4x)
The graph
2(4cosπ−41)
=
Use the examples entering the upper and lower limits of integration.