Integral of 12ctg3x 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:
∫12cot(3x)dx=12∫cot(3x)dx
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Rewrite the integrand:
cot(3x)=sin(3x)cos(3x)
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There are multiple ways to do this integral.
Method #1
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Let u=sin(3x).
Then let du=3cos(3x)dx and substitute 3du:
∫3u1du
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The integral of a constant times a function is the constant times the integral of the function:
∫u1du=3∫u1du
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The integral of u1 is log(u).
So, the result is: 3log(u)
Now substitute u back in:
3log(sin(3x))
Method #2
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Let u=3x.
Then let du=3dx and substitute 3du:
∫3sin(u)cos(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)cos(u)du=3∫sin(u)cos(u)du
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Let u=sin(u).
Then let du=cos(u)du and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(sin(u))
So, the result is: 3log(sin(u))
Now substitute u back in:
3log(sin(3x))
So, the result is: 4log(sin(3x))
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Add the constant of integration:
4log(sin(3x))+constant
The answer is:
4log(sin(3x))+constant
The answer (Indefinite)
[src]
/
|
| 12*cot(3*x) dx = C + 4*log(sin(3*x))
|
/
∫12cot(3x)dx=C+4log(sin(3x))
The graph
/ /3*pi\\
-4*log|sin|----||
\ \ 16 //
−4log(sin(163π))
=
/ /3*pi\\
-4*log|sin|----||
\ \ 16 //
−4log(sin(163π))
Use the examples entering the upper and lower limits of integration.