1. ## Integration by parts

Hello!

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

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

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

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

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

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

2. ## Re: Integration by parts

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

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

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

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

parts again with the last integral expression ...

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

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

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

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

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

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