Integral of (sin4x+1)dx dx
The solution
Detail solution
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Integrate term-by-term:
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Let u=4x.
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)
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
The result is: x−4cos(4x)
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Add the constant of integration:
x−4cos(4x)+constant
The answer is:
x−4cos(4x)+constant
The answer (Indefinite)
[src]
/
| cos(4*x)
| (sin(4*x) + 1) dx = C + x - --------
| 4
/
∫(sin(4x)+1)dx=C+x−4cos(4x)
The graph
45−4cos(4)
=
45−4cos(4)
Use the examples entering the upper and lower limits of integration.