To integrate the function int cos(2x) dx using the method of integration by parts, we start by recalling the integration by parts formula:
∫u dv = uv – ∫v du
Here, we will need to choose parts of the integral. Let’s set:
- u = cos(2x) –> this implies du = -2sin(2x)dx
- dv = dx –> this gives v = x
Now we can apply the formula:
∫cos(2x) dx = x * cos(2x) – ∫(x)(-2sin(2x)) dx
This can be simplified to:
∫cos(2x) dx = x * cos(2x) + 2∫x * sin(2x) dx
Next, we need to tackle the new integral ∫x * sin(2x) dx, which also requires integration by parts. Let:
- u = x –> then du = dx
- dv = sin(2x) dx –> leading to v = -½cos(2x)
Applying the integration by parts formula here provides:
∫x * sin(2x) dx = x * (-½cos(2x)) – ∫(-½cos(2x)) dx
This simplifies to:
∫x * sin(2x) dx = -½x * cos(2x) + ¼sin(2x) + C
Substituting this back into our previous expression:
∫cos(2x) dx = x * cos(2x) + 2(-½x * cos(2x) + ¼sin(2x) + C)
This simplifies to:
∫cos(2x) dx = x * cos(2x) – x * cos(2x) + ½sin(2x) + C
Thus, the integral ∫cos(2x) dx evaluates to:
∫cos(2x) dx = ¼sin(2x) + C
In conclusion, the integral of cos(2x) using integration by parts leads to a final result of ¼sin(2x) + C.