Integral of (x^2)*cos(3x) dx
The solution
Detail solution
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x2 and let dv(x)=cos(3x).
Then du(x)=2x.
To find v(x):
-
Let u=3x.
Then let du=3dx and substitute 3du:
∫3cos(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=3∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 3sin(u)
Now substitute u back in:
3sin(3x)
Now evaluate the sub-integral.
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=32x and let dv(x)=sin(3x).
Then du(x)=32.
To find v(x):
-
Let u=3x.
Then let du=3dx and substitute 3du:
∫3sin(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫sin(u)du=3∫sin(u)du
-
The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −3cos(u)
Now substitute u back in:
−3cos(3x)
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫(−92cos(3x))dx=−92∫cos(3x)dx
-
Let u=3x.
Then let du=3dx and substitute 3du:
∫3cos(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)du=3∫cos(u)du
-
The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 3sin(u)
Now substitute u back in:
3sin(3x)
So, the result is: −272sin(3x)
-
Add the constant of integration:
3x2sin(3x)+92xcos(3x)−272sin(3x)+constant
The answer is:
3x2sin(3x)+92xcos(3x)−272sin(3x)+constant
The answer (Indefinite)
[src]
/
| 2
| 2 2*sin(3*x) x *sin(3*x) 2*x*cos(3*x)
| x *cos(3*x) dx = C - ---------- + ----------- + ------------
| 27 3 9
/
∫x2cos(3x)dx=C+3x2sin(3x)+92xcos(3x)−272sin(3x)
The graph
2*cos(3) 7*sin(3)
-------- + --------
9 27
92cos(3)+277sin(3)
=
2*cos(3) 7*sin(3)
-------- + --------
9 27
92cos(3)+277sin(3)
Use the examples entering the upper and lower limits of integration.