Results 1 to 13 of 13

Thread: anti deriviative

  1. #1
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64

    anti deriviative

    the question is :
    anti deriviative of : 4/(1+4t^2)
    How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    Quote Originally Posted by frankinaround View Post
    the question is :
    anti deriviative of : 4/(1+4t^2)
    How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
    antiderivative will be an arctan function
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by frankinaround View Post
    the question is :
    anti deriviative of : 4/(1+4t^2)
    How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
    Product and chain rules are used for differentiation NOT anti-differentiation.

    To do this particular anti-differentiation you make the substitution $\displaystyle 2t = \tan(\theta)$.

    Alternatively you can substitute $\displaystyle u = 2t$ and recognise a standard form.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    but how do you know when its integration by parts or trig substitution?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    With practice it becomes clearer. You can either learn the trig forms or try integration by parts and when that fails badly it's a trig sub
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
    theyta = arctan(2t)
    t=1/2 (because its an area problem from t= 0 to t = 1/2)

    The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
    -pie
    3/2 * pie
    1/2 * pie
    pie
    0

    How do I get the right answer? How do I calculate the arctan without a calculator? ect?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    $\displaystyle \displaystyle \int_0^{\frac{1}{2}} \frac{4}{1+4t^2} \, dt $

    $\displaystyle u = 2t$ ... $\displaystyle du = 2 \, dt$

    $\displaystyle \displaystyle 2\int_0^{\frac{1}{2}} \frac{2}{1+(2t)^2} \, dt$

    $\displaystyle \left[2\arctan(2t)\right]_0^{\frac{1}{2}}$

    $\displaystyle 2\arctan(1) - 2\arctan(0)$ ... note the following:

    (1) these are unit circle values. you should already be familiar with them.
    (2) the arctangent function has range $\displaystyle -\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ ... you should be familiar with this also.

    $\displaystyle 2 \cdot \dfrac{\pi}{4} - 2(0) = \dfrac{\pi}{2}$
    Last edited by skeeter; Apr 3rd 2011 at 11:04 AM. Reason: fix a dt
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,119
    Thanks
    718
    Awards
    1
    Quote Originally Posted by frankinaround View Post
    for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
    theyta = arctan(2t)
    t=1/2 (because its an area problem from t= 0 to t = 1/2)

    The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
    -pie
    3/2 * pie
    1/2 * pie
    pie
    0

    How do I get the right answer? How do I calculate the arctan without a calculator? ect?
    If you have arctan(1) then ask yourself the question for what values of theta is tan(theta) = 1?

    By the way, it's "theta" not "theyta" and "pi" not "pie."

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :

    In = integral.

    In(4/(1+4t^2))
    4t^2=tan^2theta
    2t=tan(Theta)
    dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta

    in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
    in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
    in(2costheta + cos(theta)sin(theta) dTheta)
    = 2 sin theta + 1/2 sin^2 theta.

    evaluate at 1/2 and 0.
    theta = arctan 2t. arctan 1 = 45* or pie/4
    2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.


    Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.

    right now to solve integrals I have these methods :
    Trig substitution
    u substitution
    factoring
    and sometimes its the equation of a circle.
    so what am i missing? arctan substitution? what?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    note the following basic rule for taking the derivative of $\displaystyle y = \arctan{u}$ where $\displaystyle u$ is a function of $\displaystyle x$ ...

    $\displaystyle \dfrac{d}{dx} \arctan{u} = \dfrac{u'}{1+u^2}$

    now, what is the antiderivative of $\displaystyle \dfrac{u'}{1+u^2}$ ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by frankinaround View Post
    so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :

    In = integral.

    In(4/(1+4t^2))
    4t^2=tan^2theta
    2t=tan(Theta)
    dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta

    in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
    in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
    in(2costheta + cos(theta)sin(theta) dTheta)
    = 2 sin theta + 1/2 sin^2 theta.

    evaluate at 1/2 and 0.
    theta = arctan 2t. arctan 1 = 45* or pie/4
    2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.


    Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.

    right now to solve integrals I have these methods :
    Trig substitution
    u substitution
    factoring
    and sometimes its the equation of a circle.
    so what am i missing? arctan substitution? what?
    If you want to use trigonometric substitution, then as was suggested, you let $\displaystyle \displaystyle t = \frac{1}{2}\tan{\theta}$ (or $\displaystyle \displaystyle 2t = \tan{\theta}$) so that $\displaystyle \displaystyle dt = \frac{1}{2}\sec^2{\theta}\,d\theta$.

    Then your integral $\displaystyle \displaystyle \int{\frac{4\,dt}{1 + 4t^2}}$ becomes

    $\displaystyle \displaystyle \int{\frac{4\cdot \frac{1}{2}\sec^2{\theta}\,d\theta}{1 + 4\left(\frac{1}{2}\tan{\theta}\right)^2}}$

    $\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + 4\cdot \frac{1}{4}\tan^2{\theta}}}$

    $\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + \tan^2{\theta}}}$

    $\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{\sec^2{\theta }}}$

    $\displaystyle \displaystyle = \int{2\,d\theta}$

    $\displaystyle \displaystyle = 2\theta + C$.


    Now remembering that $\displaystyle \displaystyle 2t = \tan{\theta}$, this means $\displaystyle \displaystyle \theta = \arctan{\left(2t\right)}$.

    So $\displaystyle \displaystyle 2\theta + C = 2\arctan{\left(2t\right)} + C$.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    but how do you make the substitution t for 1/2 tan theta?

    from what I understand :

    sqrt(1-x^2) can be a trig substitution of a triangle with a hypotenuse of 1 and a side x and another side sqrt(1-x^2)

    So what is the triangle with 1/2tan theta = t? can you help me see the triangle this is all coming from?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    I don't use a triangle AT ALL. The point is to turn the denominator into either $\displaystyle \displaystyle 1 - \sin^2{\theta}$ or $\displaystyle \displaystyle 1 - \tan^2{\theta}$ so that you can simplify everything using a trigonometric identitity. You can substitute ANY function you like that will make this happen, as long as you substitute its derivative for $\displaystyle \displaystyle dx$ as well.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 1st and 2nd deriviative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Nov 21st 2010, 06:52 PM
  2. Deriviative help.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 6th 2010, 08:23 AM
  3. Replies: 2
    Last Post: May 10th 2010, 02:17 AM
  4. Deriviative of function #4
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Aug 4th 2009, 11:45 AM
  5. deriviative using log rules
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 10th 2009, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum