Integral of 2*sin(3x) 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(3x)dx=2∫sin(3x)dx
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Let u=3x.
Then let du=3dx and substitute 3du:
∫3sin(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=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)
So, the result is: −32cos(3x)
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Add the constant of integration:
−32cos(3x)+constant
The answer is:
−32cos(3x)+constant
The answer (Indefinite)
[src]
/
| 2*cos(3*x)
| 2*sin(3*x) dx = C - ----------
| 3
/
∫2sin(3x)dx=C−32cos(3x)
The graph
Use the examples entering the upper and lower limits of integration.