Integral of (1+sin4x)/(sin2x)^2 dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
sin2(2x)sin(4x)+1=sin2(2x)sin(4x)+sin2(2x)1
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Integrate term-by-term:
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Rewrite the integrand:
sin2(2x)sin(4x)=−cos(x)2sin(x)+sin(x)cos(x)1
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫(−cos(x)2sin(x))dx=−2∫cos(x)sin(x)dx
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Let u=cos(x).
Then let du=−sin(x)dx 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(x))
So, the result is: 2log(cos(x))
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Don't know the steps in finding this integral.
But the integral is
−2log(sin2(x)−1)+log(sin(x))
The result is: −2log(sin2(x)−1)+log(sin(x))+2log(cos(x))
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Don't know the steps in finding this integral.
But the integral is
−2sin(2x)cos(2x)
The result is: −2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2sin(2x)cos(2x)
Method #2
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Rewrite the integrand:
sin2(2x)sin(4x)+1=−cos(x)2sin(x)+sin(x)cos(x)1+4sin2(x)cos2(x)1
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Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−cos(x)2sin(x))dx=−2∫cos(x)sin(x)dx
-
Let u=cos(x).
Then let du=−sin(x)dx and substitute −du:
∫(−u1)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫u1du=−∫u1du
-
The integral of u1 is log(u).
So, the result is: −log(u)
Now substitute u back in:
−log(cos(x))
So, the result is: 2log(cos(x))
-
Don't know the steps in finding this integral.
But the integral is
−2log(sin2(x)−1)+log(sin(x))
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The integral of a constant times a function is the constant times the integral of the function:
∫4sin2(x)cos2(x)1dx=4∫sin2(x)cos2(x)1dx
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Don't know the steps in finding this integral.
But the integral is
−sin(2x)2cos(2x)
So, the result is: −2sin(2x)cos(2x)
The result is: −2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2sin(2x)cos(2x)
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Now simplify:
−2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2tan(2x)1
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Add the constant of integration:
−2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2tan(2x)1+constant
The answer is:
−2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2tan(2x)1+constant
The answer (Indefinite)
[src]
/
| / 2 \
| 1 + sin(4*x) log\-1 + sin (x)/ cos(2*x)
| ------------ dx = C + 2*log(cos(x)) - ----------------- - ---------- + log(sin(x))
| 2 2 2*sin(2*x)
| sin (2*x)
|
/
∫sin2(2x)sin(4x)+1dx=C−2log(sin2(x)−1)+log(sin(x))+2log(cos(x))−2sin(2x)cos(2x)
∞−2iπ
=
∞−2iπ
Use the examples entering the upper and lower limits of integration.