Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - difficult integral

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    148

    difficult integral

    Hi everyone,

    Can someone give me a hint on this one, please?

    I have the integral from 1 to the square root of 3 arctan(1/x) dx

    I said that u=arctan(1/x)
    du=????
    dv=???????
    v=???????

    I'm not sure if my u is correct or not. I don't think it is because now I can't choose anything for du, right? Should it be u=arctan and du=1/x????

    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \displaystyle u=\arctan\frac{1}{x}
    \displaystyle du=-\frac{1}{1+x^2}
    dv=1
    v=x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by chocolatelover View Post
    I have the integral from 1 to the square root of 3 arctan(1/x) dx

    I said that u=arctan(1/x)
    du=????
    dv=???????
    v=???????
    First, you need to get the antiderivative of \int\arctan\frac1x\,dx

    So you apply integration by parts.

    Take u=\arctan\frac1x. Now, to get du, remember that (\arctan y)'=\frac1{1+y^2}y' (chain rule); and take dv=dx\implies v=x

    Try it, tell us later.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2007
    Posts
    148
    This is what I got:

    u=arctan(1/x)
    du=1/1+x^2dx
    dv=dx
    v=x

    integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx=

    square root of 3tan^-1(1-sqauare root of 3)-[1(tan^-1)(1-1)]-integral 1 to the square root of three x/1+x^2dx=

    -1.095-0-integral 1 to the square root of three x/1+x^2dx

    -1.095-the integral 0 to the square root of 9x/1+x^2dx

    t=1+x^2
    dt=2xdx
    xdx=1/2dt

    when x=square root of 3 t=4
    when x=1 t=2

    integral 1 to square root of 3 x/1+x^2dx=1/2 integral 2 to 4 dt/t=1/2ln(t)]2 to 4=
    1/2(ln4-ln2)=1/2ln2=

    integral 1 to square root of 3 tan^-1xdx=-1.095-integral 1 to square root 3 x/1+x^2=
    -1.095-1/2ln2

    Can you tell me if this is correct and if it isn't where I made my mistake?

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by chocolatelover View Post
    u=arctan(1/x)
    du=1/1+x^2dx
    Correct this.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    148
    u=arctan(1/x)
    du=1/1+x^2dx

    Correct this.


    u=arctan(1/x)
    du=[1/1+x^2]ln(x)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chocolatelover View Post
    This is what I got:

    u=arctan(1/x)
    du=1/1+x^2dx
    dv=dx
    v=x

    integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx=

    square root of 3tan^-1(1-sqauare root of 3)-[1(tan^-1)(1-1)]-integral 1 to the square root of three x/1+x^2dx=

    -1.095-0-integral 1 to the square root of three x/1+x^2dx

    -1.095-the integral 0 to the square root of 9x/1+x^2dx

    t=1+x^2
    dt=2xdx
    xdx=1/2dt

    when x=square root of 3 t=4
    when x=1 t=2

    integral 1 to square root of 3 x/1+x^2dx=1/2 integral 2 to 4 dt/t=1/2ln(t)]2 to 4=
    1/2(ln4-ln2)=1/2ln2=

    integral 1 to square root of 3 tan^-1xdx=-1.095-integral 1 to square root 3 x/1+x^2=
    -1.095-1/2ln2

    Can you tell me if this is correct and if it isn't where I made my mistake?

    Thank you very much
    by the way, chocolatelover, writing out sentences to describe mathematical symbols and such makes your post tedious to read, at least for me. if you can't use LaTex, try these shorthands, it will make your posts more readable AND save you a lot of typing

    instead of: the square root of x
    type: sqrt(x)

    instead of: the integral of xdx
    type: int(x)dx

    instead of: the integral from a to b of xdx
    type: int_{a}^{b} (x)dx

    OR

    int (x)dx ...from a to b (keep the limits at the end of the line after you type everything else)

    example:

    instead of typing, as you did: integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx

    you could type: int_{1}^{sqrt(3)}arctan(1/x)dx = xarctan(1/x) - int_{1}^{sqrt(3)}[x/(1+x^2)]dx

    it is understandable that the limits for xarctan(1/x) is from 1 to sqrt(3), if you want to emphasis it, leave off the limits to the end

    int [arctan(1/x)]dx = xarctan(1/x) - int [x/(1+x^2)]dx ...from 1 to sqrt(3)


    now, see how nicer and shorter that is?

    Or better yet, you could learn to use LaTex, it's not as hard as you may think. See our LaTex Tutorial
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chocolatelover View Post

    u=arctan(1/x)
    du=[1/1+x^2]ln(x)
    as red_dog and Krizalid said.

    u = arctan(1/x)

    => du = -1/(1 + x^2)


    why is that?

    By the chain rule: \frac {d}{dx} \arctan u = \frac {1}{1 + u^2}u'

    So, \frac {d}{dx} \arctan \left( \frac {1}{x} \right) = \frac {1}{1 + \left( \frac {1}{x} \right)^2} \cdot \frac {-1}{x^2}

    ............................ = - \frac {1}{1 + x^2}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2007
    Posts
    148
    Is this correct so far?

    u=arctan(1/x)
    du=-1/1+x^2
    dv=dx
    v=x

    =tan^-1(x/)(-1/1+x^2)]1-sq. root 3-integr.1-squ.3-1/1+x^2dx
    =tan^-1(1/sq.3)(-1/1+ sq. 3^2-tan^-1(1/1)(-1/1+1^2)-integ. 1-sq.3 -1/1+x^2dx
    =-.131-.785-integ.1-sq.3-1/1+x^2dx

    Thank you
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chocolatelover View Post
    Is this correct so far?

    u=arctan(1/x)
    du=-1/1+x^2
    dv=dx
    v=x

    =tan^-1(x/)(-1/1+x^2)]1-sq. root 3-integr.1-squ.3-1/1+x^2dx
    =tan^-1(1/sq.3)(-1/1+ sq. 3^2-tan^-1(1/1)(-1/1+1^2)-integ. 1-sq.3 -1/1+x^2dx
    =-.131-.785-integ.1-sq.3-1/1+x^2dx

    Thank you
    no, that's not correct. and what you typed makes no sense, did you read through the last post i made?

    Here's how to do it.

    Using
    u = \arctan \left( \frac {1}{x} \right) \implies du = - \frac {1}{1 + x^2}

    dv = dx \implies v = x

    We proceed using integration by parts

    \int_{1}^{\sqrt {3}} \arctan \left( \frac {1}{x} \right)~dx = \left. x \arctan \left( \frac {1}{x} \right) \right|_{1}^{\sqrt {3}} + \int_{1}^{\sqrt {3}} \frac {x}{1 + x^2}~dx

    for the last integral, we use the substitution: u = 1 + x^2

    \Rightarrow \int_{1}^{\sqrt {3}} \arctan \left( \frac {1}{x} \right)~dx = \left. x \arctan \left( \frac {1}{x} \right) + \frac {1}{2} \ln (1 + x^2) \right|_{1}^{\sqrt {3}}

    I leave plugging in the limits to you
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2007
    Posts
    148
    Hi,

    "arctan" is the same thing as tan^-1 right?

    Can you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx

    How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?

    Where you said "we use the substitution u: 1+x^2" is du=x
    dv=2x
    v=x^2/2

    To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?

    Thank you
    Last edited by chocolatelover; September 2nd 2007 at 11:17 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    again, you put the wrong slash on the [tex] tag! please follow instructions carefully when we type them, we don't do it just for fun, we do it to help you
    Quote Originally Posted by chocolatelover View Post
    Hi,

    "arctan" is the same thing as tan^-1 right?
    yes


    Can you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx
    the formula has a minus sign in it, yes, but du is negative, a negative times a negative is positive

    How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?
    The formula for integration by parts is:

    \int u~dv = uv - \int v~du

    u = \arctan \left( \frac {1}{x} \right) and v = x

    thus uv = x \arctan \left( \frac {1}{x} \right)


    Where you said "we use the substitution u: 1+x^2" is du=x
    dv=2x
    v=x^2/2
    no. integration by substitution is different from integration by parts, there is no need to find dv and v when doing substitution

    To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?

    Thank you
    you plug in \sqrt {3} first and then minus the answer you get when you plug in 1

    this is what the Fundamental Theorem of Calculus says. It's a very important theorem, please look it up in your text, you must know it
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Aug 2007
    Posts
    148
    you plug in first and then minus the answer you get when you plug in 1
    In other words, sqr. 3arctan(1/sqr3)+1/2ln(1+sqr. 3^2)-[1arctan(1/1)+1/2ln(1+1^2) right? My answer will be a decimal number, right? I won't have pie or anything like that, right?

    Thank you
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chocolatelover View Post
    In other words, sqr. 3arctan(1/sqr3)+1/2ln(1+sqr. 3^2)-[1arctan(1/1)+1/2ln(1+1^2) right? My answer will be a decimal number, right? I won't have pie or anything like that, right?

    Thank you
    yes, it will be a decimal if you evaluate it. but you can leave the answer "exact" if you want. meaning no decimals. just work out the arctan's and simplify the ln's
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Aug 2007
    Posts
    148
    just work out the arctan's and simplify the ln's
    You can't simplify this any farther, can you?

    Sqr. 3 acr tan(1/sqr. 3)- arctan(1)+1/2ln(4)-1/2ln(1)

    If you can simplify it, can you show me how to?

    Thank you
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Difficult integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 15th 2010, 12:33 PM
  2. Difficult integral!?!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 20th 2010, 11:33 AM
  3. Difficult Integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 16th 2010, 03:20 PM
  4. Difficult Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 11:08 AM
  5. Difficult integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 30th 2008, 10:27 AM

Search Tags


/mathhelpforum @mathhelpforum