Integral of sin^4 dx
The solution
Detail solution
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Rewrite the integrand:
sin4(x)=(21−2cos(2x))2
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
(21−2cos(2x))2=4cos2(2x)−2cos(2x)+41
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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:
∫4cos2(2x)dx=4∫cos2(2x)dx
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Rewrite the integrand:
cos2(2x)=2cos(4x)+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(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)
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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)
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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:
∫2cos(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=2∫cos(u)du
-
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)
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The integral of a constant is the constant times the variable of integration:
∫41dx=4x
The result is: 83x−4sin(2x)+32sin(4x)
Method #2
-
Rewrite the integrand:
(21−2cos(2x))2=4cos2(2x)−2cos(2x)+41
-
Integrate term-by-term:
-
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 integral of a constant times a function is the constant times the integral of the function:
∫(−2cos(2x))dx=−2∫cos(2x)dx
-
Let u=2x.
Then let du=2dx and substitute 2du:
∫2cos(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=2∫cos(u)du
-
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 integral of a constant is the constant times the variable of integration:
∫41dx=4x
The result is: 83x−4sin(2x)+32sin(4x)
-
Add the constant of integration:
83x−4sin(2x)+32sin(4x)+constant
The answer is:
83x−4sin(2x)+32sin(4x)+constant
The answer (Indefinite)
[src]
/
|
| 4 sin(2*x) sin(4*x) 3*x
| sin (x) dx = C - -------- + -------- + ---
| 4 32 8
/
∫sin4(x)dx=C+83x−4sin(2x)+32sin(4x)
The graph
3
3 3*cos(1)*sin(1) sin (1)*cos(1)
- - --------------- - --------------
8 8 4
−83sin(1)cos(1)−4sin3(1)cos(1)+83
=
3
3 3*cos(1)*sin(1) sin (1)*cos(1)
- - --------------- - --------------
8 8 4
−83sin(1)cos(1)−4sin3(1)cos(1)+83
3/8 - 3*cos(1)*sin(1)/8 - sin(1)^3*cos(1)/4
Use the examples entering the upper and lower limits of integration.