- #1

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## 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

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- Thread starter jpd5184
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- #1

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

none

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

- #2

tiny-tim

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(have a a square-root: √ and try using the X

i don't understand how you got 2/3(1/sqrt(1-u^2)) from (2/(sqrt.(1-t^2))dt

- #3

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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

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first i have to make sure … if using substitution is the right method for solving the problem. is it?

substitution

but i doubt that u = t

try getting it right, and see

- #5

Mark44

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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 = t

- #6

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[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 = t^{3}, du = 3t^{2}. 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

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No, du= 3t^2 dt## 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, 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.1/3 du = t^2

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

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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

Mark44

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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

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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

Mark44

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

- #12

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pi/12 isnt giving me the right answer

- #13

Mark44

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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]

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