Integral of (cos3x-2x^3) dx
The solution
Detail solution
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−2x3)dx=−2∫x3dx
-
The integral of xn is n+1xn+1 when n=−1:
∫x3dx=4x4
So, the result is: −2x4
-
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)
The result is: −2x4+3sin(3x)
-
Add the constant of integration:
−2x4+3sin(3x)+constant
The answer is:
−2x4+3sin(3x)+constant
The answer (Indefinite)
[src]
/
| 4
| / 3\ x sin(3*x)
| \cos(3*x) - 2*x / dx = C - -- + --------
| 2 3
/
∫(−2x3+cos(3x))dx=C−2x4+3sin(3x)
The graph
1 sin(3)
- - + ------
2 3
−21+3sin(3)
=
1 sin(3)
- - + ------
2 3
−21+3sin(3)
Use the examples entering the upper and lower limits of integration.