- #1
- 35
- 0
Cyclovenom said:Let's review what you got
[tex] 2x = \sec\theta [/tex]
[tex] \sqrt{4x^2-1}= tan\theta [/tex]
[tex] dx = \frac{sec\theta tan\theta d\theta}{2} [/tex]
so you will have:
[tex] \int \frac{(\frac{sec\theta}{2})^3}{tan\theta}\frac{sec\theta tan\theta d\theta}{2} [/tex]
and working the terms:
[tex] \frac{1}{16} \int sec^4\theta d\theta [/tex]
[tex] \frac{1}{16} \int sec^2\theta (1+tan^2\theta) d\theta [/tex]
[tex] \frac{1}{16} \int sec^2\theta d\theta + sec^2\theta tan^2\theta d\theta [/tex]
[tex] \frac{1}{16} [\int sec^2\theta d\theta + \int sec^2\theta tan^2\theta d\theta] [/tex]
and you can finish it yourself, i believe...
marlon said:[tex] [\int sec^2\theta tan^2\theta d\theta] [/tex]
is equal to tan³(theta)/3
regards
marlon
Cyclovenom said:On your triangle you should have on:
Hypotenuse: [tex] 2x [/tex]
Opposite: [tex] \sqrt{4x^2-1} [/tex]
Adjacent:[tex] 1 [/tex]
HallsofIvy said:Well, not "u= 2 sec θ" because there was no "u" in the integral!
But setting 2x= secθ will work very nicely (Notice where the 2 is!)
sin^{2}θ+ cos^{2}θ= 1 so, dividing both sides by cos^{2}θ, tan^{2}θ+ 1= sec^{2}θ and then
sec^{2}θ- 1= tan^{2}θ.
That's the whole point of the substitution. If 2x= sec θ then 4x^{2} =
sec^{2} θ so 4x^{2}- 1= tan^{2}θ and you can drop the square root: [itex]\sqrt{4x^2-1}= \sqrt{tan^2\theta}= tan \theta [/itex].
(You might have been thinking of examples involving "x^{2}- 4" where the substitution x= 2sec θ would give 4sec^{2}- 4= 4(tan^{2} θ)