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Improper integral help (trig sub?)

  • #1
I have been bashing my head against this problem for a couple of hours now and cannot for the life of me figure it out. i am able to get AN answer but when i check it with my calculator i always get pi

Homework Statement



integral from (0, 1) of:

(4r*dr)/sqrt(1 - r^4)

Homework Equations



The Attempt at a Solution


i know the equation is undefined at r = 1 so:

lim t --> 1 of the integral (0, t) of (4r*dr)/sqrt(1 - r^4)

by trig subsitution:
sin(x) = sqrt(1-r^4)
cos(x) = r^2
deriving implicitly gives:
-sin(x) = 2rdr
-2sin(x) = 4rdr

so i end up with:
integral (0, t) of (-2sin(x))/(sin(x))

which gives me the answer as -2


maybe my calculator is wrong? or more likely im making some stupid mistake/overlooking something.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
128
0
When you do a variable substitution you need to change the limits of integration as well.
So you need to integrate from the x-value when [tex] \sin(x) = 1[/tex] to where [tex]\sin(x) = 0 [/tex] since when [tex]r = 0, \quad \sqrt{1-r^4} = 1[/tex] and when [tex] r = 1, \quad \sqrt{1-r^4}= 0[/tex].

[tex] \pi [/tex] is the correct answer.
 
Last edited:
  • #3
612
1
try r = [cos(θ)]^1/2

also with your sub i get r^4 = cos(x)
 
  • #4
128
0
try r = [cos(θ)]^1/2

also with your sub i get r^4 = cos(x)
His substitution is fine. He just forgot to change the limits.
Also [tex] \sin(x) = \sqrt{1-\cos^2 (x)} [/tex] so in his case [tex] r^4 = \cos^2 (x) \Rightarrow r^2 = \cos(x) [/tex]
 
  • #5
When you do a variable substitution you need to change the limits of integration as well.
[tex] \pi [/tex] is the correct answer. So you need to integrate from the x-value when sin(x) = 1 to where sin(x) = 0 since when r = 0 sqrt(1-r^4) = 1 and when r = 1 sqrt (1-r^4) = 0.
yup, youre right i was making a stupid mistake, i just totally left out the last step.

i just plugged arccos(r^2) in for theta after integrating and got pi as the correct answer. thanks for all of your help.
 
  • #6
612
1
His substitution is fine. He just forgot to change the limits.
Also [tex] \sin(x) = \sqrt{1-\cos^2 (x)} [/tex] so in his case [tex] r^4 = \cos^2 (x) \Rightarrow r^2 = \cos(x) [/tex]
Was just pointing out there is an easier sub where he may have not run into arithmetic mistakes. Like his claim: sin(x) = sqrt(1-r^4) -> cos(x) = r^2
 
  • #7
128
0
Was just pointing out there is an easier sub where he may have not run into arithmetic mistakes. Like his claim: sin(x) = sqrt(1-r^4) -> cos(x) = r^2
What is wrong with that claim? Looks right to me.


Edit:
Think a right angled triangle with hypotenuse 1 and the opposite side of the angle x as sqrt(1-r^4).
This means that sin(x) = sqrt(1-r^4).
From Pythagoras the last side (y) must satisfy sqrt(1-r^4)^2 + y^2 = 1^2 --> 1 - r^4 + y^2 = 1 --> r^4 = y^2 --> y = r^2.
Now cos(x) = y/1 = r^2.
 
Last edited:
  • #8
612
1
ah oops i was doing sin(x) = sqrt(1-r^4) ⇒ sin^2(X) = 1 – r^4 ⇒ r^4 = sin^2(x) – 1 ⇒ r^4 = cos^2(x)

and just noticed that he didn't have a ^2 on his cos(x) my bad.
 

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