Integration by parts

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August 13th 2012, 03:36 PM
Lotte1990

Integration by parts

Hello!

I am trying to solve the following indefinite integral by using integration by parts (as is recommended by the textbook): $\int{(e^x\sin{x})dx}$ .

However, I am stuck in the same circle over and over again.

$U = e^x$ and $dU = (e^x)dx$ .
$dV = (\sin{x})dx$ and $V = -\cos{x}$ .

So: $\int{(U)dV} = UV - \int{(V)dU}$ gives $\int{(e^x\sin{x})dx} = -e^x\cos{x} - \int{(-e^x\cos{x})dx}$ .

Resulting in the same problem when trying to find $\int{(-e^x\cos{x})dx}$ .

Even when I define $U$ and $dV$ the other way around, I am still stuck with $e^x$ and either $\sin{x}$ or $\cos{x}$ .

How do I do this? Please help!
August 13th 2012, 03:50 PM
skeeter

Re: Integration by parts

$\int e^x \sin{x} \, dx$

$u = \sin{x} \, , \, dv = e^x \, dx$

$du = \cos{x} \, dx \, , \, v = e^x$

$\int e^x \sin{x} \, dx = e^x \sin{x} - \int e^x \cos{x} \, dx$

parts again with the last integral expression ...

$u = \cos{x} \, , \, dv = e^x \, dx$

$du = -\sin{x} \, dx \, , \, v = e^x$

$\int e^x \sin{x} \, dx = e^x \sin{x} - \left[e^x \cos{x} - \int e^x (-\sin{x}) \. dx \right]$

$\int e^x \sin{x} \, dx = e^x \sin{x} - e^x \cos{x} - \int e^x \sin{x} \, dx$

$2\int e^x \sin{x} \, dx = e^x \sin{x} - e^x \cos{x}$

$\int e^x \sin{x} \, dx = \frac{e^x(\sin{x}-\cos{x})}{2} + C$

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