Integral of cos^2t*sin^2t dt
The solution
Detail solution
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Rewrite the integrand:
sin2(t)cos2(t)=(21−2cos(2t))(2cos(2t)+21)
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There are multiple ways to do this integral.
Method #1
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Let u=2t.
Then let du=2dt 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:
8t−32sin(4t)
Method #2
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Rewrite the integrand:
(21−2cos(2t))(2cos(2t)+21)=41−4cos2(2t)
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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:
∫41dt=4t
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The integral of a constant times a function is the constant times the integral of the function:
∫(−4cos2(2t))dt=−4∫cos2(2t)dt
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Rewrite the integrand:
cos2(2t)=2cos(4t)+21
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫2cos(4t)dt=2∫cos(4t)dt
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Let u=4t.
Then let du=4dt 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(4t)
So, the result is: 8sin(4t)
-
The integral of a constant is the constant times the variable of integration:
∫21dt=2t
The result is: 2t+8sin(4t)
So, the result is: −8t−32sin(4t)
The result is: 8t−32sin(4t)
Method #3
-
Rewrite the integrand:
(21−2cos(2t))(2cos(2t)+21)=41−4cos2(2t)
-
Integrate term-by-term:
-
The integral of a constant is the constant times the variable of integration:
∫41dt=4t
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−4cos2(2t))dt=−4∫cos2(2t)dt
-
Rewrite the integrand:
cos2(2t)=2cos(4t)+21
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫2cos(4t)dt=2∫cos(4t)dt
-
Let u=4t.
Then let du=4dt 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(4t)
So, the result is: 8sin(4t)
-
The integral of a constant is the constant times the variable of integration:
∫21dt=2t
The result is: 2t+8sin(4t)
So, the result is: −8t−32sin(4t)
The result is: 8t−32sin(4t)
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Add the constant of integration:
8t−32sin(4t)+constant
The answer is:
8t−32sin(4t)+constant
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.