Integral of cos(x)+sin(3x) dx
The solution
Detail solution
-
Integrate term-by-term:
-
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)
-
The integral of cosine is sine:
∫cos(x)dx=sin(x)
The result is: sin(x)−3cos(3x)
-
Add the constant of integration:
sin(x)−3cos(3x)+constant
The answer is:
sin(x)−3cos(3x)+constant
The answer (Indefinite)
[src]
/
| cos(3*x)
| (cos(x) + sin(3*x)) dx = C - -------- + sin(x)
| 3
/
∫(sin(3x)+cos(x))dx=C+sin(x)−3cos(3x)
The graph
Use the examples entering the upper and lower limits of integration.