Integration of ∫e^(-1/2)t cos t dt

**crabchef** · June 29th 2010, 11:08 AM

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

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 $u=\cos t$ and $dv=e^{(-1/2)t} dv$ , cos just turns into sin x, which turns into -cos x, etc and e^(-1/2)t never goes away either.

Please help!

**skeeter** · June 29th 2010, 11:28 AM

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

$\int e^{kt} \cos{t} \, dt<br />$

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

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

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

parts again w/ the last integral term ...

$u = \sin{t}$ ... $dv = e^{kt} \, dt<br />$

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

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

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

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

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

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

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

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

**crabchef** · June 29th 2010, 03:13 PM

Thanks a bunch skeeter

I would have never figured that one out =\

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