Integrating 3/(9+4x^2)

glenn101 · 2009-06-19 19:26:48

I need help doing this integration, I don't quite understand how it works, I do use an approach for these types of questions but I just keep getting lost in manipulating the integral. I attempted to solve it via this approach:

3
∫ ----------
9+4x^2

= 3/2
2∫ --------------
4(9/4+x^2)
3
= 1/6∫ ----------------
2(9/4+x^2)
3/2
= 1/6∫ ---------------
9/4+x^2

= 1/6Tan^-1(1/3x/2)+c

= 1/6Tan^-1(2x/3)+c

however the answer says its: 1/2Tan^-1(2x/3)+c

What am I doing wrong??? I'm so confused:(

Last edited by glenn101 (2009-06-19 22:16:48)

bobbym · 2009-06-19 19:52:47

Hi glen101

Your 3rd line is incorrect. 2 is okay.

glenn101 · 2009-06-19 19:55:21

How do I go about fixing it? thats what I don't understand, how is it incorrect?

Last edited by glenn101 (2009-06-19 19:58:14)

bobbym · 2009-06-19 20:00:56

Hi glenn101;

Line 2 = 1/2Tan^-1(2x/3)+c

Line 3 = 1/6Tan^-1(2x/3)+c

Whatever you did to go from 2 to 3 is off by a factor of 1/3

glenn101 · 2009-06-19 20:03:09

hmm I see what you mean, I just can't find a definate way of getting rid of that 4 on the denominator, thats what I'm trying to achieve. I'm getting completely confused now, I've been at this the whole day:(

I'll show you another question which I used the same approach, this one worked however.
1
∫ ----------------
9+16t^2
1
= ∫ ------------------
16(9/16+t^2)
3/4
= 4/3∫ ----------------
16(9/16+t^2
12
= 1/12∫ --------------
16(9/16+t^2)
3
= 1/12∫ ----------------
4(9/16+t^2)
3/4
= 1/12∫-----------------
9/16+t^2

= 1/12Tan^-1(4/3t)+c

Could someone show me how they would go about integrating this:|

Last edited by glenn101 (2009-06-19 20:14:28)

bobbym · 2009-06-19 21:02:45

Hi glenn101;

To get rid of the 1/4 just pull it out of the integral like any other constant and bring it to the left of the integral operator.

This is how I would do that integral.

Try to get the integral into this form

#1

Start:

now you have it in the form that is the sum of 2 squares with a=3/2. Now just plug into #1

Last edited by bobbym (2009-06-22 09:11:04)

glenn101 · 2009-06-19 21:53:53

Thanks so much bobbym, I'm new to integration, I got confused whether you can take values to the left side of the integrand. I was unaware that it worked with every constant. My teacher unfortunately never explained to us those little side notes, so me and my classmates have been left to find these out for ourselves.
Again many thanks, I can get a good rest tonight knowing how this works.

Oh and one last thing, In my text book, it has it as the form:

a
----------
a^2+x^2

not 1
---------
a^2+x^2

which is correct?

Last edited by glenn101 (2009-06-19 21:56:42)

bobbym · 2009-06-19 21:57:27

Hi glenn101;

#1 is for 1 in the numerator. Just pull the a to the left side of the integral and continue as I have shown.

I looked at this page and it goes much further than I did here. Fills in some of the missing pieces of my notes.

http://en.wikipedia.org/wiki/Trigonometric_substitution

Last edited by bobbym (2009-06-19 22:00:13)

glenn101 · 2009-06-19 22:04:34

Thanks for the additional information bobbym.
I think your form works under the conditions you stated, mine presented in my textbook is as follows:
a
∫ ------------ = Tan^-1(x/a)+c
a^2+x^2

Last edited by glenn101 (2009-06-19 22:09:57)

bobbym · 2009-06-19 22:13:52

Thats odd I checked three other sources and they agree with

#1

with a>0

Last edited by bobbym (2009-06-19 22:14:47)

glenn101 · 2009-06-19 22:16:13

hmmm, similarly with sine and cosine:
1
∫ --------------- =sin^-1(x/a)+c
√a^2-b^2

-1
∫ --------------- =cos^-1(x/a)+c
√a^2-b^2

My textbook is:
Cambridge Essential Specialist Mathematics Third Edition

Last edited by glenn101 (2009-06-19 22:17:46)

bobbym · 2009-06-19 22:28:18

From a Short Table of Integrals by Pierce (Integral 48) gives it as #1. I don't know why the discrepancy. Oh , I see it ! The form with a in the numerator cancels the 1/a in my form. Both are right.

Last edited by bobbym (2009-06-19 22:30:32)

glenn101 · 2009-06-19 22:49:14

Ah, so either works. Thanks for verifying that. I think I best get on practicing these problems. I can't thank you enough for your efforts bobbym.

Last edited by glenn101 (2009-06-19 22:49:23)

Math Is Fun Forum

#1 2009-06-19 19:26:48

Integrating 3/(9+4x^2)

#2 2009-06-19 19:52:47

Re: Integrating 3/(9+4x^2)

#3 2009-06-19 19:55:21

Re: Integrating 3/(9+4x^2)

#4 2009-06-19 20:00:56

Re: Integrating 3/(9+4x^2)

#5 2009-06-19 20:03:09

Re: Integrating 3/(9+4x^2)

#6 2009-06-19 21:02:45

Re: Integrating 3/(9+4x^2)

#7 2009-06-19 21:53:53

Re: Integrating 3/(9+4x^2)

#8 2009-06-19 21:57:27

Re: Integrating 3/(9+4x^2)

#9 2009-06-19 22:04:34

Re: Integrating 3/(9+4x^2)

#10 2009-06-19 22:13:52

Re: Integrating 3/(9+4x^2)

#11 2009-06-19 22:16:13

Re: Integrating 3/(9+4x^2)

#12 2009-06-19 22:28:18

Re: Integrating 3/(9+4x^2)

#13 2009-06-19 22:49:14

Re: Integrating 3/(9+4x^2)

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