Let
![I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cint%20e%5E%7B-2x%7D%20%5Ccos%282x%29%20%5C%2C%20dx%5B%2F%5Dtex%3C%2Fp%3E%3Cp%3EIntegrate%20by%20parts%3A%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdisplaystyle%20%5Cint%20u%20%5C%2C%20dv%20%3D%20uv%20-%20%5Cint%20v%20%5C%2C%20du)
with

Then

Integrate by parts again, this time with

so that

Answer:

Step-by-step explanation:
is in the third quadrant
To find the reference angle subtract π from it, that is
reference angle =
- π = 
Its1. every number/variable that has a power to the zero, its goin to turn out to be just one
Answer: 0.36
Step-by-step explanation:
6/10 x 6.10= 36/100
Answer:
x=-5
Step-by-step explanation:
-2x-41=5x-6
-35=7x
-5=x