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Math Help - 2 short calc problems

  1. #1
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    2 short calc problems

    1. Verify the following identity:
    <br />
\ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|<br />

    2. Find the antiderivative:
    <br />
\int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt<br />

    I really don't understand these two, need some help
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  2. #2
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    Quote Originally Posted by nirva
    1. Verify the following identity:
    <br />
\ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|<br />
    Note if,
    f'(x)=g'(x)
    Then,
    f(x)=g(x)+C for sum real number C.
    This is the method which I presume your teacher wants you to prove this.

    I am going to find the derivatives of these two function. Notice that,
    (\ln |x|)'=1/x
    (\csc x)'=-\csc x\cot x
    (\cot x)'=-\csc^2x
    ---------
    Thus given,
    y=\ln| \csc x-\cot x|
    Then by the chain rule you have,
    y'=(\csc x-\cot x)'\cdot \frac{1}{\csc x-\cot x}
    Thus,
    y'=\frac{(-\csc x\cot x +\csc^2x)}{\csc x-\cot x}=\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}=\csc x

    Next given,
    y=-\ln|\csc x+\cot x|
    Then,
    y'=-(\csc x+\cot x)'\cdot \frac{1}{\csc x+\cot x}
    Thus,
    y'=\frac{(\csc x\cot x+\csc^2x)}{\csc x+\cot x}=\frac{\csc x(\cot x+\csc x)}{\csc x+\cot x}=\csc x

    Since their derivative match implies that the function differ by a constant. Thus,
    \ln |\csc x-\cot x|=-\ln|\csc x+\cot x|+C
    Let x=\pi/4, then,
    \csc x=\sqrt{2}
    \cot x=1
    Thus, substituting these values,
    \ln|\sqrt{2}-1|=-\ln|1+\sqrt{2}|+C
    Note that,
    \sqrt{2}-1>0,1+\sqrt{2}>0 thus,
    \ln (\sqrt{2}-1)=-\ln (\sqrt{2}+1)+C
    Call n=\sqrt{2}-1 then, n^{-1}=\sqrt{2}+1, then,
    \ln n=-\ln n^{-1}+C
    Using the rule of exponents for logarithms we have,
    \ln n=-(-1)\ln n+C
    Thus,
    \ln n=\ln n+C
    Thus, C=0
    Thus,
    \ln |\csc x-\cot x|=-\ln|\csc x+\cot x|

    Note: The reason why if n=\sqrt{2}-1 then, n^{-1}=\sqrt{2}+1 is simple. Because,
    n^{-1}=\frac{1}{\sqrt{2}-1}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1
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  3. #3
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    Quote Originally Posted by nirva
    2. Find the antiderivative:
    <br />
\int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt<br />

    I really don't understand these two, need some help
    Let,
    x=\3t^2-4 then,
    2x+13=6t^2+5 and,
    dx/dt=6t
    Thus, by substitution,
    \int \cos (2x+13)\cos x \frac{dx}{dt}dt=\int \cos (2x+13)\cos xdx
    Using the Identity,
    \cos x\cos y=(1/2)[\cos (x+y)+\cos (x-y)]
    We have,
    \frac{1}{2}\int \cos (3x+13) +\cos (x+13) dx
    This is a linear substitution function henceforth,
    \frac{1}{2}\cdot \left( \frac{1}{3}\sin (3x+13)+\sin (x+13) +C_1\right)
    Thus,
    \frac{1}{6} \sin (3x+13)+\frac{1}{2} \sin (x+13)+C
    Substituting back,
    \frac{1}{6}\sin (3t^2+1)+\frac{1}{2}\sin (t^2+9)+C
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  4. #4
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    Quote Originally Posted by nirva
    1. Verify the following identity:
    <br />
\ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|<br />

    2. Find the antiderivative:
    <br />
\int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt<br />

    I really don't understand these two, need some help
    Here are some ways.

    A) Prove:
    -ln|cscX +cotX| = ln|cscX -cotX|

    Reminders:
    --log(a/b) = log(a) -log(b)
    --log(1/b) = log(1) -log(b) = 0 -log(b) = -log(b)

    So, -ln|cscX +cotX| == ln|1/(cscX +cotX)| ------**

    Trig Identity:
    sin^2(X) +cos^2(X) = 1 -------------**
    Divide both sides by sin^2(X),
    1 +cot^2(X) = csc^2(X)
    Rearrange,
    1 = csc^2(X) -cot^2(X)
    Factor the RHS, [a^2 -b^2 == (a+b)(a-b)],
    1 = (cscX +cotX)(cscX -cotX)
    Divide both sides by (cscX +cotX),
    1/(cscX +cotX) = cscX -cotX
    Take the natural logs of the absolute values of both sides,
    ln|1/(cscX +cotX)| = ln|cscX -cotX|
    Simplifying,
    -ln|cscX +cotX| = ln|cscX -cotX|

    Therefore, proven.

    ==================================
    2) Find Antiderivative of
    6t*cos(6t^2 +5)cos(3t^2 -4)dt --------(i)

    Here, if we can separate the two cosines, we are in good shape.

    Trig Identity:
    cosAcosB = (1/2)[cos(A-B) +cos(A+B)] ------**

    So, in (i), if A = (6t^2 +5) and B = (3t^2 -4),
    cos(6t^2 +5)cos(3t^2 -4)
    = (1/2)[cos(3t^2 +9) +cos(9t^2 +1)]
    And (i) becomes,
    (1/2)[cos(3t^2 +9) +cos(9t^2 +1)](6t dt) ----(ia)

    Now it is easy to find the antiderivative of that (ia).

    INT.{(1/2)[cos(3t^2 +9) +cos(9t^2 +1)](6t dt)}
    = (1/2)INT.[cos(3t^2 +9) +cos(9t^2 +1)](6t dt) ----(ii)

    Let u = 3t^2 +9, so du = 6t dt
    And V = 9t^2 +1, so dV = 18t dt

    For simplicity, let us expand the (ii).
    (You do not have to do this if you like. This is just for easy explanation.)
    = (1/2)INT.[cos(3t^2 +9)](6t dt) +(1/2)INT.[cos(9t^2 +1)](6t dt)
    Then, easy sailing,
    = (1/2)sin(3t^2 +5) +(1/2)INT.[cos(9t^2 +1)](18t)/3 dt
    = (1/2)sin(3t^2 +5) +(1/3)(1/2)INT.[cos(9t^2 +1)](18t dt)
    = (1/2)sin(3t^2 +5) +(1/6)sin(9t^2 +1) -------------------answer.

    For your exercise, try finding the derivative of that answer. It should match (i).

    -------------------
    For your other postings, if you can post only one or two problems per posting, I may try to solve them, if I have time.
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