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Math Help - Trigonometric Calculus

  1. #1
    Zel is offline
    Newbie Zel's Avatar
    Feb 2010

    Question Trigonometric Calculus

    I am having a bit of trouble with these, so if anyone could help me, I would greatly appreciate it.
    Thanks in advance

    1. Find from first principles the derivative of the function f(x) = tan(2x)

    a) Find the area between the graph of y = tan x and the x axis for -1≤ x ≤ 1, giving your answer to two decimal places
    b) Using calculus, show that ∫ tan x dx = -ln| cos x| + c, where c is an arbitrary constant
    c) Find a number k, such that ∫ tan x dx = 2, for 0≤ x≤ k, giving your answer to two decimal places

    Thanks again
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  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    2. b) \int{\tan{x}\,dx} = \int{\frac{\sin{x}}{\cos{x}}\,dx}

     = -\int{\left(\frac{1}{\cos{x}}\right)(-\sin{x})\,dx}.

    Now let u = \cos{x} so that \frac{du}{dx} = -\sin{x} and the integral becomes


     = -\int{\frac{1}{u}\,du}

     = -\ln{|u|} + C

     = -\ln{|\cos{x}|} + C.
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  3. #3
    MHF Contributor

    Apr 2005
    For 1, I presume you mean using \lim_{h\to 0}\frac{tan(x+h)- tan(x)}{h}.

    tan(x)= \frac{sin(x)}{cos(x)} so that is \lim_{h\to 0}\frac{\frac{sin(x+h)}{cos(x+h)}- \frac{sin(x)}{cos(x)}}{h} =\lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x+h)}{h cos(x)cos(x+h)} = \lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x)+ sin(x)cos(x)- cos(x)sin(x+ h)}{h cos(x)cos(x+h)} = \left(cos(x)\lim(h\to 0}\frac{sin(x+h)- sin(x)}{h}- cos(x+h)\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}\right)\left(\lim_{h\to 0}\frac{1}{cos(x)cos(x+h)}

    Use limits you know about sine and cosine.

    Prove It showed how to do 2b. That should make 2a easy. For 2c, set the formula for \int_0^k tan(x)dx, that you get from 2b, equal to 2 and solve for k.
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