Integral of sen^2xcos^2xdx dx
The solution
Detail solution
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Rewrite the integrand:
sin2(x)cos2(x)=(21−2cos(2x))(2cos(2x)+21)
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There are multiple ways to do this integral.
Method #1
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Let u=2x.
Then let du=2dx and substitute du:
∫(81−8cos2(u))du
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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:
∫81du=8u
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The integral of a constant times a function is the constant times the integral of the function:
∫(−8cos2(u))du=−8∫cos2(u)du
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Rewrite the integrand:
cos2(u)=2cos(2u)+21
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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:
∫2cos(2u)du=2∫cos(2u)du
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Let u=2u.
Then let du=2du and substitute 2du:
∫2cos(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=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(2u)
So, the result is: 4sin(2u)
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The integral of a constant is the constant times the variable of integration:
∫21du=2u
The result is: 2u+4sin(2u)
So, the result is: −16u−32sin(2u)
The result is: 16u−32sin(2u)
Now substitute u back in:
8x−32sin(4x)
Method #2
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Rewrite the integrand:
(21−2cos(2x))(2cos(2x)+21)=41−4cos2(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:
∫41dx=4x
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The integral of a constant times a function is the constant times the integral of the function:
∫(−4cos2(2x))dx=−4∫cos2(2x)dx
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Rewrite the integrand:
cos2(2x)=2cos(4x)+21
-
Integrate term-by-term:
-
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
-
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)
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The integral of a constant is the constant times the variable of integration:
∫21dx=2x
The result is: 2x+8sin(4x)
So, the result is: −8x−32sin(4x)
The result is: 8x−32sin(4x)
Method #3
-
Rewrite the integrand:
(21−2cos(2x))(2cos(2x)+21)=41−4cos2(2x)
-
Integrate term-by-term:
-
The integral of a constant is the constant times the variable of integration:
∫41dx=4x
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−4cos2(2x))dx=−4∫cos2(2x)dx
-
Rewrite the integrand:
cos2(2x)=2cos(4x)+21
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫2cos(4x)dx=2∫cos(4x)dx
-
Let u=4x.
Then let du=4dx and substitute 4du:
∫4cos(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=4∫cos(u)du
-
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 integral of a constant is the constant times the variable of integration:
∫21dx=2x
The result is: 2x+8sin(4x)
So, the result is: −8x−32sin(4x)
The result is: 8x−32sin(4x)
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Add the constant of integration:
8x−32sin(4x)+constant
The answer is:
8x−32sin(4x)+constant
The answer (Indefinite)
[src]
/
|
| 2 2 sin(4*x) x
| sin (x)*cos (x) dx = C - -------- + -
| 32 8
/
∫sin2(x)cos2(x)dx=C+8x−32sin(4x)
The graph
1 cos(2)*sin(2)
- - -------------
8 16
−16sin(2)cos(2)+81
=
1 cos(2)*sin(2)
- - -------------
8 16
−16sin(2)cos(2)+81
Use the examples entering the upper and lower limits of integration.