Integral of sin(2x)^2 dx
The solution
Detail solution
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Rewrite the integrand:
sin2(2x)=21−2cos(4x)
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫21dx=2x
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2cos(4x))dx=−2∫cos(4x)dx
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Let u=4x.
Then let du=4dx and substitute 4du:
∫4cos(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=4∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 4sin(u)
Now substitute u back in:
4sin(4x)
So, the result is: −8sin(4x)
The result is: 2x−8sin(4x)
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Add the constant of integration:
2x−8sin(4x)+constant
The answer is:
2x−8sin(4x)+constant
The answer (Indefinite)
[src]
/
|
| 2 x sin(4*x)
| sin (2*x) dx = C + - - --------
| 2 8
/
∫sin2(2x)dx=C+2x−8sin(4x)
The graph
1 cos(2)*sin(2)
- - -------------
2 4
−4sin(2)cos(2)+21
=
1 cos(2)*sin(2)
- - -------------
2 4
−4sin(2)cos(2)+21
Use the examples entering the upper and lower limits of integration.