# Thread: Integration of ∫e^(-1/2)t cos t dt

1. ## Integration of ∫e^(-1/2)t cos t dt

How would you integrate $\displaystyle \int e^{(-1/2)t}(\cos t)\,dt$?

I tried integrating by parts, but the problem I discovered is that it doesn't work because when you integrate either function, it just yields another function. For example, if I choose $\displaystyle u=\cos t$ and $\displaystyle dv=e^{(-1/2)t} dv$, cos just turns into sin x, which turns into -cos x, etc and e^(-1/2)t never goes away either.

2. let $\displaystyle k = -\frac{1}{2}$

$\displaystyle \int e^{kt} \cos{t} \, dt$

$\displaystyle u = \cos{t}$ ... $\displaystyle dv = e^{kt} \, dt$

$\displaystyle du = -\sin{t} \, dt$ ... $\displaystyle v = \frac{1}{k}e^{kt}$

$\displaystyle \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k} \int e^{kt} \sin{t} \, dt$

parts again w/ the last integral term ...

$\displaystyle u = \sin{t}$ ... $\displaystyle dv = e^{kt} \, dt$

$\displaystyle du = \cos{t} \, dt$ ... $\displaystyle v = \frac{1}{k}e^{kt}$

$\displaystyle \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k}\left[\frac{1}{k}e^{kt} \sin{t} - \frac{1}{k} \int e^{kt} \cos{t} \, dt \right]$

$\displaystyle \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k^2}e^{kt} \sin{t} - \frac{1}{k^2} \int e^{kt} \cos{t} \, dt$

$\displaystyle \int e^{kt} \cos{t} \, dt + \frac{1}{k^2} \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k^2}e^{kt} \sin{t}$

back substitute $\displaystyle k = -\frac{1}{2}$ ...

$\displaystyle \int e^{-\frac{1}{2}t} \cos{t} \, dt + 4\int e^{-\frac{1}{2}t} \cos{t} \, dt = -2e^{-\frac{1}{2}t} \cos{t} + 4e^{-\frac{1}{2}t} \sin{t}$

$\displaystyle 5\int e^{-\frac{1}{2}t} \cos{t} \, dt = -2e^{-\frac{1}{2}t} \cos{t} + 4e^{-\frac{1}{2}t} \sin{t}$

$\displaystyle \int e^{-\frac{1}{2}t} \cos{t} \, dt = -\frac{2}{5}e^{-\frac{1}{2}t} \cos{t} + \frac{4}{5}e^{-\frac{1}{2}t} \sin{t} + C$

3. Thanks a bunch skeeter I would have never figured that one out =\