How can I simplify these integration problems using trigonometric substitutions?

[Odyssey]

(1) \alpha(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^ 2-u^2}}

(2) \beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^ 2+b}}

(3) \beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^ 2-b}}

Should I use trig subs? If so, what should my "u" be?

 

[trancefishy]

i've not done integrals with the limits and equalities such as you have posted, but, the definite integrals are easy enough.

use the trig identities sin^2 x + cos^2 x = 1

for the first problem, let a = u sin x, da = u cos x
from there, you can see that you can factor out a u^2 in the radical sign. you are left with sin^2 x - 1, which equals cos^2 x. the square root of cos^2 x is cos x.
now you should have definite integral of 1/(sin^2 x cos x)
use power reduction to simpliy sin^2 x in terms of cos. i got to go to class now, sorry. if nobody has gotten to it in 4 hours from now, i'll be back. and also learn the tex commands so this is readable

 

[ReyChiquito]

for the first one use u=asin(x) or u=acos(x)
second one u=b^{1/2}tan(x)
third u=b^{1/2}sec(x)

you must be very carefull with the sign of the functions thoug, remember that
(x^2)^{1/2}=|x|

 

[Odyssey]

thank you!

 

[HallsofIvy]

In general, if you see \srqt{1- x^2} you should immediately think "cos²= 1- sin²".

If you see \sqrt{1+ x^2} you should immediately think "1+ tan²= sec²".

If you see \sqrt{x^2- 1} you should immediately think "sec²- 1= tan²".