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