Integral of 2*cos(3*x) 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:
∫2cos(3x)dx=2∫cos(3x)dx
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Let u=3x.
Then let du=3dx and substitute 3du:
∫3cos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=3∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 3sin(u)
Now substitute u back in:
3sin(3x)
So, the result is: 32sin(3x)
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Add the constant of integration:
32sin(3x)+constant
The answer is:
32sin(3x)+constant
The answer (Indefinite)
[src]
/
| 2*sin(3*x)
| 2*cos(3*x) dx = C + ----------
| 3
/
∫2cos(3x)dx=C+32sin(3x)
/3*x\
2*sin|---|
\ 2 /
----------
3
32sin(23x)
=
/3*x\
2*sin|---|
\ 2 /
----------
3
32sin(23x)
Use the examples entering the upper and lower limits of integration.