- #1
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!)
sin2θ+ cos2θ= 1 so, dividing both sides by cos2θ, tan2θ+ 1= sec2θ and then
sec2θ- 1= tan2θ.
That's the whole point of the substitution. If 2x= sec θ then 4x2 =
sec2 θ so 4x2- 1= tan2θ 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 "x2- 4" where the substitution x= 2sec θ would give 4sec2- 4= 4(tan2 θ)
Math substitution is a method for solving equations in which one variable is replaced with an equivalent expression in terms of another variable. This allows for simplification and solving for the unknown variable.
To use math substitution, you must first identify a variable that you want to eliminate from the equation. Then, you can replace that variable with an equivalent expression in terms of another variable. You can then solve for the unknown variable using algebraic manipulation.
Math substitution can be useful for simplifying complex equations and solving for unknown variables. It also allows for the use of algebraic manipulations, which can help in understanding the relationships between different variables in an equation.
Math substitution can only be used if there is a variable that can be easily eliminated by replacing it with an equivalent expression. Some equations may not have this type of variable, making math substitution not a feasible method for solving them.
Math substitution can be used in linear equations, quadratic equations, and systems of equations. However, it may not be as effective in other types of equations, such as exponential or trigonometric equations.