Homework Help: 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

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 by a moderator: 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:
$$\frac{tan^3\theta}{3} + C$$

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

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

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.