I know this might be easy but i forgot how to solve

  • Thread starter Thread starter idiotblue
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around integration techniques in calculus, specifically focusing on the integration of functions involving trigonometric identities and substitutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants suggest various methods for integration, including integration by parts and trigonometric substitution. There is also a shift in focus to a different integral involving sine, raising questions about the calculation of cosecant and cotangent.

Discussion Status

Some participants have provided guidance on potential methods for solving the integrals, while others have shifted the discussion to different integrals, indicating a mix of interpretations and approaches. There is no explicit consensus on a single method being the best approach.

Contextual Notes

One participant notes a specific integration range from epsilon to pi minus epsilon, which may imply constraints on the problem setup. Additionally, there is mention of confusion regarding trigonometric functions, suggesting a need for clarification on definitions or calculations.

idiotblue
Messages
4
Reaction score
0
Please help me..

integrate this

\int\frac{1}{U\sqrt{1-U^2}}dU
 
Physics news on Phys.org
..Try using integration by parts and see if you can solve

u=\frac{1}{U} find du/dx
\frac{dv}{dx}=\frac{1}{\sqrt{1-U^2}} so get v=

and

\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx
 
It can be done straightforwardly with a trig substitution, like u=sin(t).
 
Sorry but...

the original question wasintegration from \epsilonto\pi-\epsilon
\int\frac{1}{sinx}dx

I tried to integrate it straight but i don't know how calculate cosec and cot...
 
Last edited:
Try the substitution u = tan(x/2).
 
:)thanks guys. I got it~
 
What is the term for the slope on an x-axis that runs parallel but never actually touches the x axis?
 
Get your own thread. But until you do 'horizontal asymptote with y value 0'.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K