# 3 Integration questions

(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?

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 gotta 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

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
$$(x^2)^{1/2}=|x|$$

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".