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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
substitution is the right way to do it,first i have to make sure … if using substitution is the right method for solving the problem. is it?
that is correct, so what i have to do is use trig substitution. ill give it a try.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^{3}, du = 3t^{2}. You omitted the dt. Omitting this will definitely cause problems in some substitutions, particularly trig substitutions.
No, du= 3t^2 dtHomework 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
so does that mean the integral of (pi/3)-(pi/4)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!