How to Calculate the Integral of 2/(sqrt(1-t^2)) dt?

[jpd5184]

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

 

[tiny-tim]

hi jpd5184!

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

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

 

[jpd5184]

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?

 

[tiny-tim]

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

 

[Mark44]

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

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.

 

[jpd5184]

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

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)

 

[HallsofIvy]

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
\int \frac{du}{u^{2/3}\sqrt{1- u^{2/3}}}
I don't think that is an improvement!

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

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

 

[jpd5184]

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

so does that mean the integral of (pi/3)-(pi/4)
just the upper limit minus the lower limit

 

[Mark44]

Fixed the upper limit of integration.

That way, the integral is
\int_{\pi/4}^{\pi/3} \frac{cos(\theta)d\theta}{cos(\theta)}= \int_{\pi/4}^{\pi/3} d\theta
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.

 

[jpd5184]

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

 

[Mark44]

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

 

[jpd5184]

pi/12 isn't giving me the right answer

 

[Mark44]

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