• Oct 17th 2006, 12:44 PM
find the integral of
f (x) = (1+x^2)ArcTan[ x]
• Oct 17th 2006, 02:10 PM
topsquark
Quote:

Originally Posted by fair_lady0072002
find the integral of
f (x) = (1+x^2)ArcTan[ x]

You'll have to be VERY careful of the domain here, but you can try

x = tan(y)
dx = sec^2(y) dy

Then 1 + x^2 --> 1 + tan^2(y) = sec^2(y)

Your integral then becomes f(y) = sec^4(y)

-Dan
• Oct 17th 2006, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by fair_lady0072002
find the integral of
f (x) = (1+x^2)ArcTan[ x]

My, where you do get these complicated integration from?

Here is what you do,

INTEGRAL (1+x^2)^2*atan(x)*(1+x^2)^{-1} dx

Let, u=atan(x) then, u'=(1+x^2)^{-1}
Then,
sec^2 u=1-x^2
cos^2 u=x^2
Then,
sec^2 u +2cos^2 u=1+x^2
Thus,
INTEGRAL (sec^2 u +2cos^2 u)^2 u*u' dx
Substitution rule it is composition of outer function,
INTEGRAL (sec^2 u+2cos^2 u)^2 du
Open parantheses, (secant*cosine=1)
INTEGRAL (sec^4 u+4cos^4 u+2) du
And this is definitely solvable
• Oct 17th 2006, 05:17 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
My, where you do get these complicated integration from?

Here is what you do,

INTEGRAL (1+x^2)^2*atan(x)*(1+x^2)^{-1} dx

Let, u=atan(x) then, u'=(1+x^2)^{-1}
Then,
sec^2 u=1-x^2
cos^2 u=x^2
Then,
sec^2 u +2cos^2 u=1+x^2
Thus,
INTEGRAL (sec^2 u +2cos^2 u)^2 u*u' dx
Substitution rule it is composition of outer function,
INTEGRAL (sec^2 u+2cos^2 u)^2 du
Open parantheses, (secant*cosine=1)
INTEGRAL (sec^4 u+4cos^4 u+2) du
And this is definitely solvable

That's a neat little trick I'll have to remember...

-Dan
• Oct 17th 2006, 06:26 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
That's a neat little trick I'll have to remember...

-Dan

When you are like me (do not use the dx,dy because their meanings have not been definied) you can only use these tricks to introduce the derivative and hence use substitution rule.

And thank you for the reputation.
• Oct 17th 2006, 07:43 PM
Soroban