Integral of y=4cosx+sin3x−8x. dx
The solution
Detail solution
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫(−8x)dx=−∫8xdx
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The integral of a constant times a function is the constant times the integral of the function:
∫8xdx=8∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: 4x2
So, the result is: −4x2
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Let u=3x.
Then let du=3dx and substitute 3du:
∫9sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫3sin(u)du=3∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −3cos(u)
Now substitute u back in:
−3cos(3x)
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The integral of a constant times a function is the constant times the integral of the function:
∫4cos(x)dx=4∫cos(x)dx
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The integral of cosine is sine:
∫cos(x)dx=sin(x)
So, the result is: 4sin(x)
The result is: −4x2+4sin(x)−3cos(3x)
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Add the constant of integration:
−4x2+4sin(x)−3cos(3x)+constant
The answer is:
−4x2+4sin(x)−3cos(3x)+constant
The answer (Indefinite)
[src]
/
| 2 cos(3*x)
| (4*cos(x) + sin(3*x) - 8*x) dx = C - 4*x + 4*sin(x) - --------
| 3
/
−3cos(3x)+4sinx−4x2
The graph
11 cos(3)
- -- + 4*sin(1) - ------
3 3
−3cos3−12sin1+11
=
11 cos(3)
- -- + 4*sin(1) - ------
3 3
−311−3cos(3)+4sin(1)
Use the examples entering the upper and lower limits of integration.