• Support PF! Buy your school textbooks, materials and every day products Here!

Integral Calculus

  • Thread starter planauts
  • Start date
  • #1
86
0

Homework Statement



Find the values of a > 0, such that [tex] \int_{a }^{a^2} \frac{1}{1+x^2} dx = 0.22[/tex]

The Attempt at a Solution



I integrated it and got:

[tex]\int_{a }^{a^2} \frac{1}{1+x^2} dx = 0.22 [/tex]

[tex]arctan(a^2) - arctan(a) = 0.22[/tex]

I am stuck after this, how would I solve this :-/

Thanks
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
Just apply <tangent> on both sides of your equation, use some trigonometry and see what you get
 
  • #3
86
0
[tex]a^2 - a - tan(0.22) = 0[/tex]
I got 1.188 and -0.188
The correct answer is a = 2.04, 2.62
 
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
You might want to look up (or prove) the correct identity for [tex]\tan (A-B)[/tex].
 
  • #5
86
0
You might want to look up (or prove) the correct identity for [tex]\tan (A-B)[/tex].
I don't think that the tangent is the problem. I think my equation is wrong for some reason. Because when I graphed it the points don't make sense.

Do you see anything wrong with my math (integrating 1/(1+x^2) etc.)?
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
The integration is ok. The trigonometry is the issue.
 
  • #7
86
0
The integration is ok. The trigonometry is the issue.
Then why doesn't the answer from the key work in the equation :-/
 
  • #8
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Then why doesn't the answer from the key work in the equation :-/
Are you using degrees instead of radians?
 
  • #9
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
Let's try it with 2.04

Equation proofing is: arctan (2.04^2) = 76.488 deg. arctan 2.04 = 63.886 deg. Difference is 13.602 degrees = 0.23 radians.

So it's correct. The other solution can be checked for accuracy in the same manner.
 
  • #10
86
0
Let's try it with 2.04

Equation proofing is: arctan (2.04^2) = 76.488 deg. arctan 2.04 = 63.886 deg. Difference is 13.602 degrees = 0.23 radians.

So it's correct. The other solution can be checked for accuracy in the same manner.
You are saying that [tex] arctan(a^2) - arctan(a) = 0.22 [/tex] is the correct equation.

So that's good. But now the problem is manipulating the equation to get the value of a.
How would I do that?

You said:
Just apply <tangent> on both sides of your equation, use some trigonometry and see what you get
I think that is where I made the mistake...I have to apply <tangent> to the whole "left hand" side as a one. I split the tangent and distributed it. I don't think that is allowed.

So if I change it: [tex] tan[arctan(a^2) - arctan(a)] = tan(0.22) [/tex]

Is this what fzero meant?
eal6jk.jpg


[tex]\frac{a^2 - a}{1 - a^2 * a} = \frac{a^2 - a}{1 - a^3} = 0.22 [/tex]

[tex]a^2 - a = 0.22 (1 - a^3)[/tex]
[tex]0 = 0.22 - 0.22 a^3 - a^2 + a[/tex]

I got a = 1 OR a = -5.3588 OR a = -0.2866
 
Last edited:
  • #11
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Is this what fzero meant?
eal6jk.jpg
Yes but

[tex]\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2}, [/tex]

so you should find

[tex]\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22 [/tex]
 
  • #12
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
Looks likes a nasty cubic one gets. Cardano's formulae should be used to solve it or Maple, Mathematica if you have them.
 
  • #13
86
0
Yes but

[tex]\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2}, [/tex]

so you should find

[tex]\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22 [/tex]
I got -0.184556093 and 1.946862606 = a, which still isn't right :(
 
  • #14
VietDao29
Homework Helper
1,423
2
Yes but

[tex]\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2}, [/tex]

so you should find

[tex]\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22 [/tex]
I got -0.184556093 and 1.946862606 = a, which still isn't right :(
There's still a little mistake there. You are taking tangent value on both sides, remember? :)

So, it should read something like this:
[tex]\tan \left( \arctan(a ^ 2) - \arctan(a) \right) = \tan (0.22)[/tex]
[tex]\Leftrightarrow \frac{a ^ 2 - a}{1 + a ^ 3} = \tan (0.22)[/tex].

Can you take it from here? :)

And, by the way, the book seems to have missed one possible value for a.
 
  • #15
86
0
There's still a little mistake there. You are taking tangent value on both sides, remember? :)

So, it should read something like this:
[tex]\tan \left( \arctan(a ^ 2) - \arctan(a) \right) = \tan (0.22)[/tex]
[tex]\Leftrightarrow \frac{a ^ 2 - a}{1 + a ^ 3} = \tan (0.22)[/tex].

Can you take it from here? :)

And, by the way, the book seems to have missed one possible value for a.
Oh my god! Thank you so much. I don't know why I wasn't seeing this...
I got 2.61737, 2.0416455 AND -0.223619421 as the roots.

Thanks to everyone who helped me, especially fzero :)
 
  • #16
VietDao29
Homework Helper
1,423
2
Oh my god! Thank you so much. I don't know why I wasn't seeing this...
I got 2.61737, 2.0416455 AND -0.223619421 as the roots.

Thanks to everyone who helped me, especially fzero :)
Yup, looks good to me.

However, since the problem just asks for positive values of a (a > 0), so you don't need to take a = -0.223619421 into consideration.
 

Related Threads on Integral Calculus

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
921
  • Last Post
Replies
5
Views
999
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
748
  • Last Post
Replies
3
Views
600
  • Last Post
Replies
2
Views
381
Top