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

Substitution nath help

  • Thread starter jpd5184
  • Start date
  • #1
76
0

Homework Statement



integral of (2/(sqrt.(1-t^2))dt evaluated at sqrt.3 / 2 and root 2 / 2.

Homework Equations



none


The Attempt at a Solution



i believe its just substition.

u=t^3
du=3t^2
1/3 du = t^2

then you get 2/3(1/sqrt(1-u^2))dt
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
hi jpd5184! :smile:

(have a a square-root: √ and try using the X2 icon just above the Reply box :wink:)

i don't understand how you got 2/3(1/sqrt(1-u^2)) from (2/(sqrt.(1-t^2))dt :confused:
 
  • #3
76
0


lets see:

from the original equation:
u= t^3
du= 3t^2
(1/3)du=t^2

ok so i messed up. i accidently substituted u in for t^2 but u=t^3.

first i have to make sure my substitution values above are right and if using substitution is the right method for solving the problem. is it?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
250
first i have to make sure … if using substitution is the right method for solving the problem. is it?
substitution is the right way to do it,

but i doubt that u = t2 or t3 is going to help …

try getting it right, and see :smile:
 
  • #5
33,503
5,191


Is this the integral?
[tex]\int_{\sqrt{2}/2}^{\sqrt{3}/2}\frac{2 dt}{\sqrt{1 - t^2}}[/tex]

It might be that you can do this one with an ordinary substitution, but this one is a natural for a trig substitution, with t = sin(u), dt = cos(u)du.

BTW, in your substitution, you had u = t3, du = 3t2. You omitted the dt. Omitting this will definitely cause problems in some substitutions, particularly trig substitutions.
 
  • #6
76
0


Is this the integral?
[tex]\int_{\sqrt{2}/2}^{\sqrt{3}/2}\frac{2 dt}{\sqrt{1 - t^2}}[/tex]

It might be that you can do this one with an ordinary substitution, but this one is a natural for a trig substitution, with t = sin(u), dt = cos(u)du.

BTW, in your substitution, you had u = t3, du = 3t2. You omitted the dt. Omitting this will definitely cause problems in some substitutions, particularly trig substitutions.
that is correct, so what i have to do is use trig substitution. ill give it a try.

lets see:

u= arcsin
du= 1/(sqrt(1-t^2)) dt

i could then take the 2 outside the integral sign and get 2(integral of arcsin)
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
955


Homework Statement



integral of (2/(sqrt.(1-t^2))dt evaluated at sqrt.3 / 2 and root 2 / 2.

Homework Equations



none


The Attempt at a Solution



i believe its just substition.

u=t^3
du=3t^2
No, du= 3t^2 dt

1/3 du = t^2
No, 1/3 du= t^2 dt and since there is no "t^2" in the numerator, you can only replace dt with 1/3 du/t^2- and you have to replace that t^2 by a function of u.

then you get 2/3(1/sqrt(1-u^2))dt[/QUOTE]
No, you don't. All you have done is put 1/3 in front and replace "t" with u.
If u= t^3, then t= u^{1/3} and t^2= u^{2/3} sqrt{1- t^2} would become sqrt{1- u^{2/3}}. Also the dt= du/t^2 becomes du/u^{2/3}. With that substitution, the integral becomes
[tex]\int \frac{du}{u^{2/3}\sqrt{1- u^{2/3}}}[/tex]
I don't think that is an improvement!

Substitution is right but much better is the substitution [itex]t= sin(\theta)[/itex]. That way, [itex]\sqrt{1- t^2}= \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta)[/itex] and [itex]dt= cos(\theta)d\theta[/itex]. Also, you should change the limits of integration from t to [itex]\theta[/itex]. The upper limit is [itex]t= \sart{3}/2= sin(\theta)[/itex] so that [itex]\theta= \pi/3[/itex]. The lower limit is [itex]t= \sqrt{2}/2= sin(\theta)[/itex] so that [itex]\theta= \pi/4[/itex].

That way, the integral is
[tex]\int_{\pi/4}^{\p/3} \frac{cos(\theta)d\theta}{cos(\theta)}= \int d\theta[/tex]
which is very easy!
 
  • #8
76
0


That way, the integral is
[tex]\int_{\pi/4}^{\p/3} \frac{cos(\theta)d\theta}{cos(\theta)}= \int d\theta[/tex]
which is very easy![/QUOTE]

so does that mean the integral of (pi/3)-(pi/4)
just the upper limit minus the lower limit
 
  • #9
33,503
5,191


Fixed the upper limit of integration.
That way, the integral is
[tex]\int_{\pi/4}^{\pi/3} \frac{cos(\theta)d\theta}{cos(\theta)}= \int_{\pi/4}^{\pi/3} d\theta[/tex]
which is very easy!
so does that mean the integral of (pi/3)-(pi/4)
just the upper limit minus the lower limit[/QUOTE]
Yes.
 
  • #10
76
0


so it would be pi/3 - pi/4
this would be 4pi/12 - 3pi/12
which would be pi/12 for the answer

since its a integral would it be pi/12 + c
 
  • #11
33,503
5,191


This is a definite integral. The answer is a specific number. For indefinite integrals, you add the constant, but not for definite integrals.
 
  • #12
76
0


pi/12 isnt giving me the right answer
 
  • #13
33,503
5,191


There is a factor of 2 that got lost along the way. The integral should be
[tex]\int_{\pi/4}^{\pi/3} \frac{2cos(\theta)d\theta}{cos(\theta)}= \int_{\pi/4}^{\pi/3} 2d\theta[/tex]
 

Related Threads on Substitution nath help

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
755
  • Last Post
2
Replies
25
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
868
Replies
3
Views
1K
Replies
21
Views
4K
Top