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