# Integrate e^2x sin x by parts [Check & Correct my working please]

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• June 8th 2011, 08:30 AM
anees7112
Integrate e^2x sin x by parts [Check & Correct my working please]
Hello

I am having trouble with integrating by parts
I am trying to integrate e^2x * sin x

I have attached my working, and I am getting stuck towards the end. Please let me know where I am going wrong.

Thank you :)
http://img709.imageshack.us/img709/3985/20110609001.jpg
• June 8th 2011, 08:36 AM
TheCoffeeMachine
It seems you've made a substitution that reversed what you did earlier:

\displaystyle \begin{aligned}\int f'(x)g(x)\;{dx} & = f(x)g(x)\;{dx}-\int f(x)g'(x)\;{dx} \\& = f(x)g(x)-f(x)g(x)+\int f'(x)g(x)\;{dx} \\& = \int f'(x)g(x)\;{dx}.\end{aligned}

The above is a pitfall to beware of when doing integration by parts.
I remember there was a similar thread a while back - see it in here.
• June 8th 2011, 08:40 AM
anees7112
Thank you, but how would I go about fixing this?
Edit: Okay, will have a look now :)
• June 8th 2011, 08:46 AM
TheCoffeeMachine
Quote:

Originally Posted by anees7112
Thank you, but how would I go about fixing this?

Let $u = e^{2x}$ again for the second integration by parts.
By the way, try this method too (the same method really).
• June 8th 2011, 08:52 AM
anees7112
Here is the problem I was having when I tried to integrate witu u as e^2x :
http://img38.imageshack.us/img38/2369/20110609002.jpg

And thanks for the link, I will look over it now
• June 8th 2011, 08:56 AM
TheCoffeeMachine
Quote:

Originally Posted by anees7112
One minute, still a small problem

Are you sure? The middle terms shouldn't cancel, and the sign of the last integral should be negative, so that you can take it to the other side.
• June 8th 2011, 09:02 AM
anees7112
I'm not sure if you saw it but I edited the last post, still stuck on something there it looks like .
• June 8th 2011, 09:05 AM
tom@ballooncalculus
Quote:

Originally Posted by TheCoffeeMachine
By the way, try this method too (the same method really).

And here's as far as the OP originally got right, in balloon sculpture.

And filling the blanks (and solving the top row for I) does confirm his/her later answer, however arrived at.

http://www.ballooncalculus.org/draw/...wice/three.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule, straight continuous lines differentiating downwards with respect to x.

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

Larger

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• June 8th 2011, 09:13 AM
anees7112
Haha I have never seen anything like that before, Tom! I would love to learn how the baloons work tonight, but I will have to postpone it until exams are over.

I tried to use wolfram to confirm (1/3)*(e^(2x)*cos x +e^(2x) sin x)=(1/5)(2 sin x- cos x) but it wouldn't give the True or False answer it sometimes does.

Anyhow, thank you again TheCoffeeMachine and Tom@balooncalculus very much :) Good night
• June 8th 2011, 09:17 AM
TheCoffeeMachine
Quote:

Originally Posted by anees7112
I'm not sure if you saw it but I edited the last post, still stuck on something there it looks like .

You forgot the minus that you had before $\cos{x}e^{2x}$ and $(2e^{2x})' = 4e^{2x}$.
• June 8th 2011, 09:23 AM
TheCoffeeMachine
Quote:

Originally Posted by tom@ballooncalculus
is lazy integration by parts

I love lazy integration by parts! (Rofl)
• June 8th 2011, 09:26 AM
anees7112
Ah okay thank you so much for taking the time to show me how to do this . One last thing, so should my second u and v' always be the same half of the equation if you know what I mean?

I let u= e^(2x) first but then for the second time I incorrectly let u=cos x instead of e^(2x) again... Is that what you meant I should be aware of when doing by parts?
• June 8th 2011, 09:38 AM
TheCoffeeMachine
Quote:

Originally Posted by anees7112
Ah okay thank you so much for taking the time to show me how to do this .

You're welcome.
Quote:

One last thing, so should my second u and v' always be the same half of the equation if you know what I mean? I let u= e^(2x) first but then for the second time I incorrectly let u=cos x instead of e^(2x) again... Is that what you meant I should be aware of when doing by parts?
Yes, basically making 'a substitution that precisely nullifies the effect of a given substitution'.
• June 8th 2011, 09:40 AM
anees7112
Great! I have spent a lot of time stuck on this question and now I can finally move on :) Night Night
• June 8th 2011, 09:44 AM
TheCoffeeMachine
Quote:

Originally Posted by anees7112
Great! I have spent a lot of time stuck on this question and now I can finally move on :) Night Night

Good night! I'm off too to celebrate my 7(Itwasntme)(Itwasntme)th post.
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