Integral of sin(3x+2)dx dx
The solution
Detail solution
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Let u=3x+2.
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+2)
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Now simplify:
−3cos(3x+2)
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Add the constant of integration:
−3cos(3x+2)+constant
The answer is:
−3cos(3x+2)+constant
The answer (Indefinite)
[src]
/
| cos(3*x + 2)
| sin(3*x + 2) dx = C - ------------
| 3
/
∫sin(3x+2)dx=C−3cos(3x+2)
The graph
cos(5) cos(2)
- ------ + ------
3 3
3cos(2)−3cos(5)
=
cos(5) cos(2)
- ------ + ------
3 3
3cos(2)−3cos(5)
Use the examples entering the upper and lower limits of integration.