Curious Question; find ⌠cotx using integration by parts

[Raza]

Homework Statement

find ⌠cotx using integration by parts with using u= 1/sinx and dv= cosx

Homework Equations

cotx=cosx/sinx

The Attempt at a Solution

u= 1/sinx and dv= cosx dx
du = -cotxcscx dx v= sinx

⌠udv = 1 + ⌠sinx cotx cscx
sinx and cscx cancel out.

⌠cotx = 1 + ⌠cotx
I=1+I
0I=1?

 

[Mark44]

Homework Statement

find ⌠cotx using integration by parts with using u= 1/sinx and dv= cosx

Homework Equations

cotx=cosx/sinx

The Attempt at a Solution

u= 1/sinx and dv= cosx dx
du = -cotxcscx dx v= sinx

⌠udv = 1 + ⌠sinx cotx cscx
sinx and cscx cancel out.

⌠cotx = 1 - ⌠cotx

Are you required to use integration by parts? If not, an ordinary substitution will work.

You have a sign error in this equation ⌠cotx = 1 - ⌠cotx . If IBP is required, try splitting u and dv differently.

 

[Raza]

Hello,
I know that it could be done with simple substitution. I was just curious of why it won't work by IBP.

Also, because I've seen ⌠tanx = ln|secx| and as -ln|cosx|

I thought that ⌠cotx = ln|sinx| would also be -ln|cscx|

was trying to prove that maybe it is also equal to -ln|cscx|.

 

[Mark44]

Don't omit the differentials and the constants of integration.

Yes, ⌠cotx dx = ln|sin x| + C = - ln(1/|csc x|) + C. And you need only the properties of logs to show that - ln(1/|csc x|) = ln|sin x|.

As I already mentioned, this is very much easier when you use substitution, but I believe that you could also use IBP if you're masochistic. I would try u = cos x and dv = csc x dx instead of what you tried.

 

[Raza]

Thank you, Mark44, I realize that it would be too hard to go through that.