Is my process is right of $\int_{}^{} e^{-2x}\cdot\cos{4x} dx $?

Question

Question : $$\int_{}^{} e^{-2x}\cdot\cos{4x} dx $$ My process : Let $$I = \int_{}^{} e^{-2x}\cdot\cos{4x} dx \quad ....[1]$$ $$I = cos{4x}\int_{}^{} e^{-2x} dx-\int_{}^{}(\frac{dx}{dt} cos4x\int_{}^{}e^{-2x}dx) dx$$ $$I = \frac{cos4x\cdot e^{-2x}}{-2} -\int_{}^{} 2e^{-2x}sin4x dx$$ $$I = -\frac{cos4x\cdot e^{-2x}}{2} -2[sin4x\int_{}^{} e^{-2x}- \int_{}^{}(\frac{dx}{dt} sin4x\int_{}^{}e^{-2x}dx) dx]$$ $$I = -\frac{cos4x\cdot e^{-2x}}{2} -2[\frac{sin4xe^{-2x}}{-2}- \int_{}^{}(4 cos4x\frac{e^{-2x}}{-2}dx) dx]$$ $$I = -\frac{cos4x\cdot e^{-2x}}{2} -e^{-2x}sin4x - 2I \quad ....from [1]$$ then , $$3I = -\frac{cos4x\cdot e^{-2x}}{2} -e^{-2x}sin4x $$ $$I = -\frac{cos4x\cdot e^{-2x}}{6} -\frac{e^{-2x}sin4x}{3} $$ Is my process is okay , i want to know that can we do it more easily .

Answer 1

The idea is correct, but there are some mistakes because you should get $$\frac{e^{-2x}(2\sin(4x)-\cos(4x))}{10}+c$$ An hint to do it more easily is to set $y:=-2x$, so you have $\text{d}x=-\frac{1}{2}\text{d}y$ and $$\int e^{-2x} \cos(4x) \text{d}x=-\frac{1}{2}\int e^{y} \cos (2y) \text{d}y$$ I have used the fact that $\cos$ is even.

Answer 2

In the last step you must get: $$I = -\frac{cos4x\cdot e^{-2x}}{2} \color{red}+e^{-2x}sin4x - \color{red}4I \quad ....from [1]$$ Hence the answer: $$I=-\frac1{10}e^{-2x}\cos 4x+\frac15\sin 4x+C.$$ Alternatively, you can guess the final answer in the form: $$\int e^{-2x}\cos 4x dx=Ae^{-2x}\cos 4x+Be^{-2x}\sin 4x+C \iff \\ (Ae^{-2x}\cos 4x+Be^{-2x}\sin 4x+C)'=e^{-2x}\cos 4x \iff \\ -2Ae^{-2x}\cos 4x-4Ae^{-2x}\sin 4x-2Be^{-2x}\sin 4x+4Be^{-2x}\cos 4x=e^{-2x}\cos 4x \iff \\ \begin{cases}-2A+4B=1\\ -4A-2B=0\end{cases} \Rightarrow A=-\frac1{10},B=\frac15.$$

Answer 3

Like this $$\int e^{-2x}\cos{4x}dx=\Re \int e^{(-2+4i)x}dx$$