# Math help

1. Sep 11, 2004

### COCoNuT

the problem is shown in the picture i attached.

wouldnt i use $$u = 2sec(\theta)$$ as a sub?

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2. Sep 11, 2004

### HallsofIvy

Staff Emeritus
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: $\sqrt{4x^2-1}= \sqrt{tan^2\theta}= tan \theta$.

(You might have been thinking of examples involving "x2- 4" where the substitution x= 2sec θ would give 4sec2- 4= 4(tan2 θ)

Last edited: Sep 11, 2004
3. Sep 11, 2004

### COCoNuT

ok so the problem should look like $$\int \frac{x^3}{tan \theta}$$ right now.

i dont know what to do next. i cant use u-du. can i use parts on this?

if you set $$2x= sec \theta$$

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

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

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

hmm my way seems longer and harder...

can you help me with your way?

Last edited: Sep 11, 2004
4. Sep 11, 2004

### Fredrik

Staff Emeritus
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

$$\frac{1}{16}\int\frac{1}{cos^4\theta}d\theta$$

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. Sep 11, 2004

### Pyrrhus

Let's review what you got

$$2x = \sec\theta$$
$$\sqrt{4x^2-1}= tan\theta$$
$$dx = \frac{sec\theta tan\theta d\theta}{2}$$

so you will have:

$$\int \frac{(\frac{sec\theta}{2})^3}{tan\theta}\frac{sec\theta tan\theta d\theta}{2}$$

and working the terms:

$$\frac{1}{16} \int sec^4\theta d\theta$$

$$\frac{1}{16} \int sec^2\theta (1+tan^2\theta) d\theta$$

$$\frac{1}{16} \int sec^2\theta d\theta + sec^2\theta tan^2\theta d\theta$$

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

and you can finish it yourself, i believe...

Last edited: Sep 11, 2004
6. Sep 11, 2004

### marlon

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. Sep 11, 2004

### COCoNuT

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

my work:
found the anti-derv.

$$\frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2$$

ok now this is the part im 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 dont know what to fill the values for the triangle.
im thinking that the values are....
h = 1/2
O = $$\sqrt{(4x^2-1)}$$

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 $$\theta$$

$$\frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2$$
will turn to...

$$\frac{1}{16}(\frac{\sqrt{(4x^2-1)}}{x^3}) + \frac{1}{16}(\frac{x^3}{2})^2$$

and that is my answer, is the correct?

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8. Sep 11, 2004

### marlon

The second integral should have solution tan³(x)/3, i think you made a mistake there.

9. Sep 11, 2004

### marlon

$$[\int sec^2\theta tan^2\theta d\theta]$$

is equal to tan³(theta)/3

regards
marlon

10. Sep 11, 2004

### COCoNuT

i dont see how you got that, i did the problem agian and got the same answer

11. Sep 11, 2004

### Pyrrhus

Marlon is right, your first integral is correct but not your second.

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

$$u=tan\theta$$
$$du=sec^2\theta d\theta$$

so we will have:
$$\int u^2 du$$

Integrating:
$$\frac{u^3}{3} + C$$

Substituting back:
$$\frac{tan^3\theta}{3} + C$$

your result should be:

$$\frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C$$

Now to change it back:

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

so
$$\frac{\sqrt{4x^2-1}}{16} + \frac{(\sqrt{4x^2-1})^3}{48} + C$$

Last edited: Sep 11, 2004
12. Sep 11, 2004

### Pyrrhus

I just noticed this is a definite integral... oh well just apply the theorem
and forget about the constant.

13. Sep 11, 2004

### COCoNuT

sorry, you guys are right....

well anyway, with the correct results...

$$\frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C$$

time to sub in theta...

$$\frac{\frac{\sqrt{(4x^2-1)}}{x^3}}{16} + (\frac{\frac{\sqrt{(4x^2-1)}}{x^3})^3}{48} + C$$

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

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

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

Last edited: Sep 11, 2004
14. Sep 11, 2004

### Pyrrhus

On your triangle you should have on:

Hypotenuse: $$2x$$
Opposite: $$\sqrt{4x^2-1}$$
Adjacent:$$1$$

15. Sep 11, 2004

### COCoNuT

sorry to keep on bothering you guys, but how did you get those values?

16. Sep 11, 2004

### Pyrrhus

Well, you started with $$2x = sec\theta$$ so if you read above HallsofIvy post on how to get a value that will work for the $$\sqrt{4x^2-1}$$ which will be $$tan\theta$$. You can build a Triangle with that info.

You know $$sec\theta = \frac{Hypotenuse}{Adjacent}$$ and you got
$$tan\theta = \frac{opposite}{adjacent}$$

so $$\frac{2x}{1} = \frac{Hypotenuse}{Adjacent}$$ and $$\frac{\sqrt{4x^2-1}}{1} = \frac{opposite}{adjacent}$$

Also:

$$\frac{1}{cos\theta} = 2x$$

$$cos\theta = \frac{1}{2x}$$

and:

$$\frac{\sqrt{4x^2-1}}{1} = \frac{sin\theta}{cos\theta}$$

$$\frac{\sqrt{4x^2-1}}{1}\frac{1}{2x} = \frac{sin\theta}{cos\theta}cos\theta$$

$$sin\theta = \frac{\sqrt{4x^2-1}}{2x}$$

$$\frac{Opposite}{Hypotenuse} = \frac{\sqrt{4x^2-1}}{2x}$$

Do you see it?

Last edited: Sep 11, 2004
17. Sep 11, 2004

### COCoNuT

yea!!! thank you, i get it.

18. Sep 11, 2004

### COCoNuT

lol, i just figured out what you tried to tell me, im a bit slow. wow, that helped alot. my teacher goes really fast, and doesnt 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. Sep 11, 2004

### Pyrrhus

Well, i'm glad i was of help
Keep solving problems!!
and feel free to ask for help anytime as long as you will learn from it.