Particular Integral of arctan example

  • #1
I know for a fact that

[tex]
\int \frac{(du)}{(a^2 + u^2)} = \frac{1}{a} \cdot arctan \frac{u}{a} + C
[/tex]

I was given the problem of solving the indefinite integral of

[tex]
\int \frac{(dx)}{(2x^2 + 2x + 5)}
[/tex]

First, I multiplied the integral by (1/2) / (1/2) to eliminate the coefficient of the x^2 in the denominator, so now I am left with

[tex]
\frac{1}{2} \cdot \int \frac{(dx)}{(x^2 + x + \frac{5}{2})}
[/tex]

Now, in completing the square of the denominator, I added [tex] \frac{1}{4} - \frac{1}{4} [/tex] (which is zero) so that the equation would look like this:

[tex]
= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{5}{2} + \frac{1}{4} - \frac{1}{4}}
[/tex]

Simplifying, I got:

[tex]
= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{1}{4} + (\frac{5}{2} - \frac{1}{4})}
[/tex]

[tex]
= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + \frac{9}{4}}
[/tex]

[tex]
= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + ( \frac{3}{2})^2}
[/tex]

If we let [tex](x + \frac{1}{2})^2 = u^2[/tex] and [tex]( \frac{3}{2})^2 = a^2[/tex] we now have the integrable form stated above, so

[tex]
= \frac{1}{2} \int \frac{du}{u^2 + a^2}
[/tex]

[tex]
= \frac{1}{2} \cdot \frac{2}{3} \cdot arctan \frac{u}{a} + C
[/tex]

[tex]
= \frac{1}{3} \cdot arctan \frac{x^2 + \frac{1}{2}}{\frac{3}{2}} + C
[/tex]

[tex]
= \frac{1}{3} \cdot arctan \frac{2(x^2 + \frac{1}{2})}{3} + C
[/tex]

My questions are:
(1) Is this the correct solution?
(2) If my solution is correct, how do you get the necessary constant in order to make the denominator of this problem a complete square? the (1/4 - 1/4) I added to the denominator just popped out of my mind. Is there any way to get it without resorting to trial and error?

thank you
 

Answers and Replies

  • #2
695
0
1) No, there seems to be some confusion regarding the subsitution of variables, you've chosen (x + 1/2)^2 = u^2, this means that u = x + 1/2, not u = (x + 1/2)^2.

2) Suppose you wanted to write x^2 + bx + c in the form (x + p)^2 + q, i.e. you want to choose p, q such that x^2 + bx + c = (x + p)^2 + q = x^2 + 2px + p^2 + q. Comparing coefficients on both sides of the equation gives 2p = b and p^2 + q = c. So you can take p = b/2 and q = c - p^2 = c - (b/2)^2.
 
Last edited:
  • #3
Zurtex
Science Advisor
Homework Helper
1,120
1
When I was doing this I found trying to complete all the quadratic equations I had by completing the square rather than any other method. Either actually solving them or just putting them in the form [itex]a(x+b)^2 + c[/itex], remember good algebra techniques greatly help calculus.
 
  • #4
Muzza said:
1) No, there seems to be some confusion regarding the subsitution of variables, you've chosen (x + 1/2)^2 = u^2, this means that u = x + 1/2, not u = (x + 1/2)^2.
um... since [tex] u^2 = ( x + \frac{1}{2} )^2 [/tex] you can directly place this in the integrable form because the form needs a [tex] u^2 [/tex],

but I understand the process of solving the quadratic eq'n or completing the square. thanx.

I was just wondering if the answer I got was correct.
 
  • #5
Zurtex
Science Advisor
Homework Helper
1,120
1
relinquished™ said:
um... since [tex] u^2 = ( x + \frac{1}{2} )^2 [/tex] you can directly place this in the integrable form because the form needs a [tex] u^2 [/tex],

but I understand the process of solving the quadratic eq'n or completing the square. thanx.

I was just wondering if the answer I got was correct.
In the final two steps you have ended up replaces u with [itex]x^2 + 1/2[/itex], that was the only mistake I think.
 
  • #6
612
1
You also picked the wrong u originally. But then when you went to substitute you put in the right one. But later when you replaced u with x you went back to the wrong u.
 
  • #7
ahh.. ok, I see the mistake. Sorry. And the title should be Particular Integral that yields arctan. My mistake as well...

the u should be

[tex] u = x + \frac{1}{2} [/tex]

and the real answer should be

[tex]= \frac{1}{3} \cdot arctan \frac{2(x + \frac{1}{2})}{3} + C[/tex]


thanx again for the clarifications and replies. ^_^
 
Last edited:
  • #8
612
1
I like your approach though; I probably would have used a ugly trig sub. I should probably memorize a few of the basic integral in tables.
 

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