# 3 Integration questions

1. Oct 4, 2004

### 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?

2. Oct 6, 2004

### 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 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

3. Oct 6, 2004

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

4. Oct 6, 2004

thank you!

5. Oct 7, 2004

### HallsofIvy

Staff Emeritus
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".