Curious Question; find ⌠cotx using integration by parts

  • Thread starter Raza
  • Start date
  • #1
203
0

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?
 
Last edited:

Answers and Replies

  • #2
35,118
6,860

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.
 
  • #3
203
0
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|.
 
  • #4
35,118
6,860
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.
 
  • #5
203
0
Thank you, Mark44, I realize that it would be too hard to go through that.
 

Related Threads on Curious Question; find ⌠cotx using integration by parts

  • Last Post
Replies
22
Views
2K
  • Last Post
Replies
5
Views
3K
Replies
4
Views
275
  • Last Post
Replies
1
Views
8K
Replies
14
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
519
  • Last Post
Replies
2
Views
2K
Top