Question

Is tan(x) integrable by parts? I know about about U-substitution, I'm interested only in integration by parts :-)

Answer 1

most likely .... sin and 1/cos are 2 different parts

Answer 2

i would say tanx most likely u sub

Answer 3

by parts means ur un winding the product rule

Answer 4

by parts is easier if u use tabular forumla only works under certain criteria

Answer 5

what is tabular formula?

Answer 6

By @amistre64's suggestion,\[\int\tan x\;dx\\
\int\sin x\sec x\;dx\]
\[\text{First try:}\;\;\begin{matrix}u=\sec x& & &dv=\sin x\;dx\\
du=\sec x\tan x\;dx& & &v=\cos x\end{matrix}\\\sec x\cos x - \int\cos x\sec x\tan x\;dx\\
1 - \int\tan x\;dx\]
This doesn't get you the known result.
\[\text{Second try:}\;\;\begin{matrix}u=\sin x& & &dv=\sec x\;dx\\
du=\cos x\;dx& & &v=\ln|\sec x+\tan x|\end{matrix}\\
\sin x\ln|\sec x+\tan x|-\int\cos x\ln|\sec x+\tan x|\;dx\]
I'm not sure if you can do anything with that...
\[\text{Third try:}\;\;\begin{matrix}u=\tan x& & &dv=dx\\
du=\sec^2x\;dx& & &v=x\end{matrix}\\
x\tan x - \int x\sec^2 x\;dx\]
Also not sure, but I haven't tried checking.

Answer 7

Yes, I've been trying these options for a few hours but get nothing. Actually, I don't want to integrate it this way, I'm just interested whether it's possible to integrate this function by parts. In definition we just need both parts to be diffirentiable and existence of antiderivative if u'v... And it seems it satisfies all conditions in the definition.

Answer 8

You could try the product of many functions method listed on the Wikipedia page for IBP. I've never learned it, so I'm not sure how to implement it right away, but that might be the way to go with this integral.

Answer 9

http://en.wikipedia.org/wiki/Integration_by_parts#Product_of_many_functions

Answer 10

thanks. I'll try