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

Having trouble with Integration by Parts

I'm real stuck with this problem of mine in The Calculus 7 by Leithold

[tex]
\int arctan \sqrt{x} dx
[/tex]

Since there is no elementary formula for integration of an inverse trigo function, we cannot manipulate the integrand in such a way as to integrate easily with one step of Integration by parts.(please verify)

So first, If we let [tex]u = arctan \sqrt{x}[/tex] and
[tex]dv = dx[/tex]

then

[tex]du = \frac{dx}{2 \sqrt{x} (x+1)}[/tex] and
[tex]v=x[/tex]

then we have

[tex]
\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}
[/tex]

Now, Case 1, without simplifying [tex]\frac{x}{\sqrt {x}}[/tex]

let [tex]db = \frac{1}{2 \sqrt{x} (x+1)}dx[/tex]
and [tex]a = x[/tex].
Then [tex] da = dx[/tex]
and [tex] b = arctan \sqrt{x} [/tex]

so

[tex]
\int arctan \sqrt{x} dx = xarctan \sqrt{x} - xarctan \sqrt{x} + \int arctan \sqrt{x} dx [/tex]

which of course will lead us nowhere but 0=0

Now, Case 2 - If we simplify [tex]\frac{x}{\sqrt {x}} = \sqrt{x}[/tex]

let [tex]a= \sqrt{x}[/tex] and [tex] db = \frac{dx}{1+x} [/tex],
then [tex]da = \frac{dx}{2 \sqrt{x}}[/tex] and [tex] b = ln (x+1)[/tex]

so

[tex]
\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \frac{1}{2} ( \sqrt{x}ln (x+1) - \int \frac{ln |1+x| dx}{2 \sqrt{x}}) [/tex]

and if we do Integration by parts again, we will just get an infinite sequence of [tex] \sqrt{x}ln (x+1) [/tex] (please verify)

I'm really open to suggestions ^_^`
 
javascript:;
 
[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}[/tex]

From there... add 1 and -1 to your numerator, and disect the expression into two... hope you figure out the next steps. (in your integral of udv)
 
0.0 hey... why didn't I think of that???

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}[/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{x + 1 - 1}{2 \sqrt{x} (x+1)}dx[/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{(x + 1)}{2 \sqrt{x} (x+1)} - \frac{1}{2 \sqrt{x} (x+1)}) dx[/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{1}{2 \sqrt{x}} - \frac{1}{2 \sqrt{x} (x+1)}) dx[/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{dx}{2 \sqrt{x}} + \int \frac{dx}{2 \sqrt{x} (x+1)}[/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{dx}{2 \sqrt{x} (x+1)})[/tex]

let [tex]u^2 = x[/tex], then [tex]u = \sqrt{x}[/tex], since [tex]du = \frac{1}{2 \sqrt{x}} [/tex]it can be integrated directly and it yields arctan...

so

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{du}{(u^2+1)} [/tex]

[tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + arctan \sqrt{x} + C[/tex]

Is this correct? ^_^;;

Man... I really don't like it when the answer is just right in your face and slaps you XD. Gotta remember the +a -a concept ^_^;;; Thanx btw Irony of Truth
 
now that's correct.
 

Related Threads for: Having trouble with Integration by Parts

  • Posted
Replies
4
Views
1K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
5
Views
6K
  • Posted
Replies
13
Views
2K
  • Posted
Replies
13
Views
2K
Replies
6
Views
3K
Replies
2
Views
5K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top