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Math Help - anti deriviative

  1. #1
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    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.
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    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
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  3. #3
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    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 2t = \tan(\theta).

    Alternatively you can substitute u = 2t and recognise a standard form.
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  4. #4
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    but how do you know when its integration by parts or trig substitution?
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    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
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    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?
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  7. #7
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    \displaystyle \int_0^{\frac{1}{2}} \frac{4}{1+4t^2} \, dt

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

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

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

    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 -\dfrac{\pi}{2} < y < \dfrac{\pi}{2} ... you should be familiar with this also.

    2 \cdot \dfrac{\pi}{4}   - 2(0) = \dfrac{\pi}{2}
    Last edited by skeeter; April 3rd 2011 at 11:04 AM. Reason: fix a dt
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    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
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  9. #9
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    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?
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  10. #10
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    note the following basic rule for taking the derivative of y = \arctan{u} where u is a function of x ...

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

    now, what is the antiderivative of \dfrac{u'}{1+u^2} ?
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  11. #11
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    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 t = \frac{1}{2}\tan{\theta} (or \displaystyle 2t = \tan{\theta}) so that \displaystyle dt = \frac{1}{2}\sec^2{\theta}\,d\theta.

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

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

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

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

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

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

    \displaystyle = 2\theta + C.


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

    So \displaystyle 2\theta + C = 2\arctan{\left(2t\right)} + C.
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  12. #12
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    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?
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  13. #13
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    I don't use a triangle AT ALL. The point is to turn the denominator into either \displaystyle 1 - \sin^2{\theta} or \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 dx as well.
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