- #1

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

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

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

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- Thread starter Odyssey
- Start date

- #1

- 87

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

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

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

- #2

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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

- 120

- 1

second one [tex]u=b^{1/2}tan(x)[/tex]

third [tex]u=b^{1/2}sec(x)[/tex]

you must be very carefull with the sign of the functions thoug, remember that

[tex](x^2)^{1/2}=|x|[/tex]

- #4

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thank you!

- #5

HallsofIvy

Science Advisor

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If you see [itex]\sqrt{1+ x^2}[/itex] you should immediately think "1+ tan

If you see [itex]\sqrt{x^2- 1}[/itex] you should immediately think "sec

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