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Calc II: integration of x arctan x

  1. Sep 21, 2004 #1


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    so, I know the integration by formula for this, since it's in my handy book of formulas, but i'm required to do this by parts, and then use a technique from the lesson to integrate it further.

    The problem: [tex] \int \,x \arctan \,x \,dx [/tex]

    (i'm assuming tan to the negative first power is the same as arctan)

    the answer: [tex] \frac{1+x^2}{2}arctan x - \frac{x}{2} [/tex]

    My process so far:

    I first looked at dividing by parts with u=x and dv= arctan x dx. That got really messy trying to integrate tan, so I flipped them over.

    What erks me is I don't feel like I'm violating any rules, I'm just getting a different answer.

    So I set u=arctan x and dv= x dx
    so [itex] du=\frac{1}{x^2+1} [/itex] and [itex] v= \frac{1}{2}x^2 [/itex]

    which leaves me with:

    (using uv - [itex] \int [/itex]dvu)

    [tex] \frac{1}{2}x^2 arctan x - \frac{1}{2}\int \frac{x^2}{x^2+1} [/tex]

    I use long division on that last term so that I have:

    [tex]\int 1 - \frac{1}{x^2+1} [/tex]

    which results in

    [tex] \frac{1}{2}x^2 arctan x - \frac{1}{2}arctan x - \frac{1}{2}x [/tex]
  2. jcsd
  3. Sep 21, 2004 #2


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    I get a postive sign for the middle term.
  4. Sep 21, 2004 #3


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    how does that help me get closer to the answer? I see my mistake of not carrying the - sign, but I'm still not quite sure how arctan adds up and substracts...
  5. Sep 21, 2004 #4

    If I understand "tan to the negative first power" correctly, and if I'm not mistaking, your assumption is not correct:
    (tan x)^-1 does not equal arctan x.
    (tan x)^-1=(sin x / cos x)^-1=cos x / sin x=cot x
  6. Sep 22, 2004 #5


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    i understood arctan as the inverse of tan, which can't be found by simply switching the fractions, that would just be cotangent, not tan to the negative first power. This is my understanding...

    I'm more concerned with how to manipulate arcsin terms so that I can get an answer that looks like the books answer.
  7. Sep 22, 2004 #6
    I agree.
    Well, if you get a positive sign for the middle term, then you have the right answer, don't you?
  8. Sep 22, 2004 #7


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    that seems to be what Tide implied.

    I manipulated it algebraically and it appears correct. But when you already know the answer, you tend to manipulate in your favor. I'll double check.

    Thanks everyone for your help
  9. Sep 23, 2004 #8
    What are you talking about with this algebraic manipulation?!?!

    You simply add like terms....

    x^2/2 + 1/2 = (1 + x^2)/2

    (remember fractions....add the numerators, keep the denominator)

    and (1/2)x is the same as x/2

    I don't see where the algebraic manipulation comes into play?
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