# Thread: Integral by parts help

1. ## Integral by parts help

I am stuck on 2 integral by parts problems.
1)$\displaystyle \int e^{ax}\sin (bx)dx$. I've been trying with both $\displaystyle u'(x)=e^{ax}$, etc. and $\displaystyle u'(x)=\sin (bx)$, etc. but didn't get far.
2)$\displaystyle \int \tan^2 (x)dx$. I obviously took $\displaystyle u'(x)=\tan (x)$ and $\displaystyle v(x)=\tan (x)$ and I've been up to $\displaystyle \int \tan ^2 (x)dx= \tan (x) \cdot ln( \cos (x))-\int ln ( \cos (x)) \cdot cosec^2(x)dx$. And stuck there.

2. Originally Posted by arbolis
I am stuck on 2 integral by parts problems.
1)$\displaystyle \int e^{ax}\sin (bx)dx$. I've been trying with both $\displaystyle u'(x)=e^{ax}$, etc. and $\displaystyle u'(x)=\sin (bx)$, etc. but didn't get far.
a similar problem is done here. i am sure you can find this exact problem lying around the forum somewhere

there is also a solution using complex numbers if you are interested. it was posted by Kriz, i am sure i can find it if need be

3. Originally Posted by arbolis
2)$\displaystyle \int \tan^2 (x)dx$. I obviously took $\displaystyle u'(x)=\tan (x)$ and $\displaystyle v(x)=\tan (x)$ and I've been up to $\displaystyle \int \tan ^2 (x)dx= \tan (x) \cdot ln( \cos (x))-\int ln ( \cos (x)) \cdot cosec^2(x)dx$. And stuck there.
i wouldn't do by parts for this guy. note that $\displaystyle \tan^2 x = \sec^2 x - 1$

4. a similar problem is done here. i am sure you can find this exact problem lying around the forum somewhere

there is also a solution using complex numbers if you are interested. it was posted by Kriz, i am sure i can find it if need be
Thanks a lot for the link! It's almost the same problem, I'll check it out more in details. Anyway, it wasn't a so easy one.
And I'll appreciate if you don't mind to spend some few time to find what did Krizalid.
And for :
i wouldn't do by parts for this guy. note that
You're right, it's quite likely not the best way to solve it, but unfortunately the problem states I have to do it with integration by parts. I'll try it harder .

5. Originally Posted by arbolis
Thanks a lot for the link! It's almost the same problem, I'll check it out more in details. Anyway, it wasn't a so easy one.
And I'll appreciate if you don't mind to spend some few time to find what did Krizalid.
And for :
if you have to do it by parts, then the post i was talking about wont help. by parts wasn't used at all. it was a bit of complex analysis

You're right, it's quite likely not the best way to solve it, but unfortunately the problem states I have to do it with integration by parts. I'll try it harder .
well, that's a pain in the patooti! (can i say patooti?) i guess i'll think about it a little later

6. Originally Posted by arbolis
You're right, it's quite likely not the best way to solve it, but unfortunately the problem states I have to do it with integration by parts. I'll try it harder .
If you have do it by parts here is the solution.

note $\displaystyle tan^2(x) = \sec x \tan x \sin x$

so you integral is $\displaystyle \int ( \sec x \tan x )(\sin x) dx$

let $\displaystyle du = \sec x \tan x$ and $\displaystyle v = \sin x$
so $\displaystyle u = \sec x$ and $\displaystyle dv = \cos x$

Bobak

7. if you have to do it by parts, then the post i was talking about wont help. by parts wasn't used at all. it was a bit of complex analysis
From what I've seen in the post, Integration by parts was used, but not exclusively. Hmm, it means the problem would get much harder if we restrain us to use only int. by parts. Maybe I should give up after thinking a bit more and after having read the whole post in your link.

8. Originally Posted by arbolis
From what I've seen in the post, Integration by parts was used, but not exclusively. Hmm, it means the problem would get much harder if we restrain us to use only int. by parts. Maybe I should give up after thinking a bit more and after having read the whole post in your link.
there seems to be a mix up, i didn't give you a link to the complex analysis post. the one i gave you is using by parts. they just have cosine instead of sine, when you switch them, it is a matter of a sign change.