Math Substitution for Solving Equations | Attached Picture Included

In summary, the problem involves finding the indefinite integral of (\sqrt{4x^2-1})/x^3 using the substitution method. After setting u= tan\theta and finding the derivative of dx, the integral simplifies to 1/16 times the integral of sec^4\theta and 1/16 times the integral of sec^2\theta tan^2\theta. Using the substitution method again, the latter integral can be simplified to 1/3 times tan^3\theta. Plugging in the values for the triangle, the result is 1/16 times the square root of 4x^2-1 plus 1/48 times the cube of the square root of 4
  • #1
COCoNuT
35
0
the problem is shown in the picture i attached.

wouldnt i use [tex]u = 2sec(\theta)[/tex] as a sub?
 

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  • #2
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)
 
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  • #3
ok so the problem should look like [tex]\int \frac{x^3}{tan \theta}[/tex] right now.

i don't know what to do next. i can't use u-du. can i use parts on this?

if you set [tex] 2x= sec \theta[/tex]

then [tex] x= \frac{sec \theta}{2}[/tex]
then would i have to take the derv. of that?
[tex]dx= 1/2*ln(sec(\theta)+tan(\theta)) [/tex]


then we would rewrite the problem as...(plugged in [tex]x=\frac{sec \theta}{2}[/tex] for x)

[tex]\int\frac{(\frac{sec \theta}{2})^3}{\sqrt{4(\frac{sec \theta}{2})^2-1}} * 1/2*ln(sec(\theta)+tan(\theta))[/tex]

hmm my way seems longer and harder...


can you help me with your way?
 
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  • #4
I'm not familiar with the notation "sec", but from what HallsofIvy wrote, I take it that sec θ is just 1/cos θ.

When I use this, and calculate dx=dx/dθ dθ, the integral simplifies to

[tex]\frac{1}{16}\int\frac{1}{cos^4\theta}d\theta[/tex]

I don't know how to solve that, but maybe you do.

The thing that you're doing wrong is the derivative. There shouldn't be any logarithms in there.
 
  • #5
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...
 
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  • #6
use the fact that 1 + tan² = 1/cos² and write the integrandum as
(1/cos²(x) * 1/cos²(x))dx.

Then you also know that d(tan(x)) = dx/cos²(x)

So you get as integrandum 1/cos²(x)*d(tan(x)) and substitute the cos² by the above expression.

regards
marlon (sorry for the writing, i really should start using LaTex)
 
  • #7
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...



[tex] \frac{1}{16} [\int sec^2\theta d\theta + \frac{1}{16}\int sec^2\theta tan^2\theta d\theta] [/tex]

my work:
found the anti-derv.



[tex] \frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2 [/tex]


ok now this is the part I am confused on... i know i need to draw a triangle and do something with it...
triangle is attached, but nothing is filled in. i took notes, but i don't know what to fill the values for the triangle.
im thinking that the values are...
a(Adjacent) = x^3
h = 1/2
O = [tex]\sqrt{(4x^2-1)}[/tex]

how do you know what the a,h, and o sides are? i just looked at my notes and copied how it worked(i should take better notes).

ok now that i filled in the triangle, now to sub it back for [tex]\theta[/tex]

[tex] \frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2 [/tex]
will turn to...


[tex] \frac{1}{16}(\frac{\sqrt{(4x^2-1)}}{x^3}) + \frac{1}{16}(\frac{x^3}{2})^2 [/tex]

and that is my answer, is the correct?
 

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  • #8
The second integral should have solution tan³(x)/3, i think you made a mistake there.
 
  • #9
[tex] [\int sec^2\theta tan^2\theta d\theta] [/tex]

is equal to tan³(theta)/3

regards
marlon
 
  • #10
marlon said:
[tex] [\int sec^2\theta tan^2\theta d\theta] [/tex]

is equal to tan³(theta)/3

regards
marlon

i don't see how you got that, i did the problem agian and got the same answer
 
  • #11
Marlon is right, your first integral is correct but not your second.

Let's work out the second integral with substitution..

[tex] u=tan\theta [/tex]
[tex] du=sec^2\theta d\theta [/tex]

so we will have:
[tex] \int u^2 du [/tex]

Integrating:
[tex] \frac{u^3}{3} + C [/tex]

Substituting back:
[tex] \frac{tan^3\theta}{3} + C [/tex]


your result should be:

[tex] \frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C [/tex]

Now to change it back:

Look at what we had above:
[tex] \sqrt{4x^2-1}= tan\theta [/tex]

so
[tex] \frac{\sqrt{4x^2-1}}{16} + \frac{(\sqrt{4x^2-1})^3}{48} + C [/tex]
 
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  • #12
I just noticed this is a definite integral... oh well just apply the theorem :smile:
and forget about the constant.
 
  • #13
sorry, you guys are right...


well anyway, with the correct results...

[tex] \frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C [/tex]


time to sub in theta...

[tex] \frac{\frac{\sqrt{(4x^2-1)}}{x^3}}{16} + (\frac{\frac{\sqrt{(4x^2-1)}}{x^3})^3}{48} + C [/tex]

is that right? i drew a triangle and everything in my eariler post


Cyclovenom got a answer of [tex] \frac{\sqrt{4x^2-1}}{16} + \frac{(\sqrt{4x^2-1})^3}{48} + C [/tex]

but i still have the x^3 in mines, what am i doing wrong? tan is o/a right? so i just subbed it in
 
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  • #14
On your triangle you should have on:

Hypotenuse: [tex] 2x [/tex]
Opposite: [tex] \sqrt{4x^2-1} [/tex]
Adjacent:[tex] 1 [/tex]
 
  • #15
Cyclovenom said:
On your triangle you should have on:

Hypotenuse: [tex] 2x [/tex]
Opposite: [tex] \sqrt{4x^2-1} [/tex]
Adjacent:[tex] 1 [/tex]


sorry to keep on bothering you guys, but how did you get those values?
 
  • #16
Well, you started with [tex] 2x = sec\theta [/tex] so if you read above HallsofIvy post on how to get a value that will work for the [tex] \sqrt{4x^2-1} [/tex] which will be [tex] tan\theta [/tex]. You can build a Triangle with that info.

You know [tex] sec\theta = \frac{Hypotenuse}{Adjacent}[/tex] and you got
[tex] tan\theta = \frac{opposite}{adjacent} [/tex]

so [tex] \frac{2x}{1} = \frac{Hypotenuse}{Adjacent}[/tex] and [tex] \frac{\sqrt{4x^2-1}}{1} = \frac{opposite}{adjacent} [/tex]

Also:

[tex] \frac{1}{cos\theta} = 2x[/tex]

[tex] cos\theta = \frac{1}{2x}[/tex]

and:

[tex]\frac{\sqrt{4x^2-1}}{1} = \frac{sin\theta}{cos\theta}[/tex]

[tex]\frac{\sqrt{4x^2-1}}{1}\frac{1}{2x} = \frac{sin\theta}{cos\theta}cos\theta[/tex]

[tex]sin\theta = \frac{\sqrt{4x^2-1}}{2x} [/tex]

[tex] \frac{Opposite}{Hypotenuse} = \frac{\sqrt{4x^2-1}}{2x} [/tex]

Do you see it?
 
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  • #17
yea! thank you, i get it.
 
  • #18
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)


lol, i just figured out what you tried to tell me, I am a bit slow. wow, that helped alot. my teacher goes really fast, and doesn't explain the little details and the book is just weird. i already solved 6 problems on my own(14 more to go) and got them all right. thank you agian, i love this forum
 
  • #19
Well, I'm glad i was of help :smile:
Keep solving problems! :cool:
and feel free to ask for help anytime as long as you will learn from it.
 

What is math substitution?

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.

How do you use math substitution to solve equations?

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.

What are the advantages of using math substitution?

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.

Are there any limitations to using math substitution?

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.

Can math substitution be used in any type of equation?

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.

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