• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Help me with my integral skills

  • Thread starter -EquinoX-
  • Start date
1. Homework Statement
[tex] \int_0^R \! \frac{2rdr}{(r^2+x^2)^{\frac{3}{2}}} [/tex]
[tex] \int_0^R \! (r^2+x^2)^{-\frac{3}{2}}}d(r^2) [/tex]

2. Homework Equations



3. The Attempt at a Solution
how did you get from line 1 to 2
 

rock.freak667

Homework Helper
6,230
31
d(r^2)=2rdr
 
oh I see... so [tex]d(r^2)[/tex] just basically means taking the derivative of [tex]r^2[/tex] with respect with r?
I am not familiar with that notation
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,476
20
Not quite. [itex]d\left(r^2\right)[/itex] is the differential of [itex]r^2[/itex] with respect to [itex]r[/itex].
 
156
0
oh I see... so [tex]d(r^2)[/tex] just basically means taking the derivative of [tex]r^2[/tex] with respect with r?
I am not familiar with that notation
I bet you are! (plus or minus a twist)

Consider [tex]I = \int (u + a^2)^{-\frac{3}{2}} \, du.[/tex]

Let [tex]u = r^2,[/tex] so [tex]du = 2r\, dr.[/tex] Then

[tex]I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}[/tex]
 
I bet you are! (plus or minus a twist)

Consider [tex]I = \int (u + a^2)^{-\frac{3}{2}} \, du.[/tex]

Let [tex]u = r^2,[/tex] so [tex]du = 2r\, dr.[/tex] Then

[tex]I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}[/tex]
thanks for clearing that up
 

Related Threads for: Help me with my integral skills

  • Posted
Replies
1
Views
1K
  • Posted
Replies
2
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top