Question

use integration by parts twice to compute \[\int e^{xy}\cos y\;dy\]

Answer 1

I have been exposed to this method in single variable calculus. can I just do this by considering the x a constant?

Answer 2

I can see that the integrand has the form \[f(x,y)=g(x,y)+h(y)\]

Answer 3

not sure if that is significant in any way.

Answer 4

do twice the integration by part, and notice that you will get same expretion. If need more help tell me

Answer 5

ok so that works for sure? I have tried it both ways and have not gotten the answer. I must be messing up somewhere in the arithmetic. if \[\int uv'=uv-\int u'v\]what is the best choice for u and v? I chose \(u=e^{xy}\) initially because it looked cleaner, but I got nowhere.

Answer 6

then I tried the opposite and still got nowhere. so I don't really know where I am going wrong!

Answer 7

You're integrating with respect to y, so treat x as a constant.
\[\begin{matrix}u=e^{xy}&dv=\cos y\;dy\\
du=xe^{xy}\;dy&v=\sin y\end{matrix}\]
For the second round of IBP, make sure you keep the same u-sub and make dv equal to the other function.

Answer 8

ok I have used exactly what you used there

Answer 9

what do you mean by using the same u-sub on the second round? that's where I am messing up!

Answer 10

keep u exponential and dv trig

Answer 11

alright I did that

Answer 12

\[\begin{align*}\int e^{xy}\cos y\;dy&=e^{xy}\sin y-\int xe^{xy}\sin y\;dy\\
&=e^{xy}\sin y-x\int e^{xy}\sin y\;dy\end{align*}\]
You'll have to use IBP again for the second integral, for which you make the subs
\[\begin{matrix}u=e^{xy}&dv=\sin y \;dy\\
du=xe^{xy}&v=-\cos y\end{matrix}\]

Answer 13

yup i'm there! well I used \(u=xe^{xy}\) but same difference.

Answer 14

what is throwing me off are the x multiplying the integrals.

Answer 15

x is just a constant now

Answer 16

your variable is y

Answer 17

right. I just have to run home from the coffee shop. battery is dying!

PS thanks for your help so far!

Also if anyone else sees this, please do not just post the answer because this is an assignment question.

Answer 18

\[\begin{align*}\color{red}{\int e^{xy}\cos y\;dy} &=e^{xy}\sin y - x\int e^{xy}\sin y\;dy\\
&= e^{xy}\sin y-x\left[-e^{xy}\cos y-\int xe^{xy}(-\cos y)\;dy\right]\\
&= e^{xy}\sin y + xe^{xy}\cos y - x^2\color{red}{\int e^{xy}\cos y\;dy}\end{align*}\]

Answer 19

so I got to this exact step. I really should have posted that to save you the latex. anyways usually I would add to both sides. in this case would I say \[(x^2+1)\int e^{xy}\cos y\; dy=e^{xy}(\sin y+x\cos y)\]

Answer 20

\[\int e^{xy}\cos y dy=e^{xy}\frac{\sin y+x\ cos y}{x^2+1}\]

Answer 21

thanks a lot BTW, I really appreciate it. all it took was the red colour!