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Thread: 2 short calc problems

  1. #1
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    2 short calc problems

    1. Verify the following identity:
    $\displaystyle
    \ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|
    $

    2. Find the antiderivative:
    $\displaystyle
    \int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt
    $

    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:
    $\displaystyle
    \ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|
    $
    Note if,
    $\displaystyle f'(x)=g'(x)$
    Then,
    $\displaystyle f(x)=g(x)+C$ for sum real number $\displaystyle C$.
    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,
    $\displaystyle (\ln |x|)'=1/x$
    $\displaystyle (\csc x)'=-\csc x\cot x$
    $\displaystyle (\cot x)'=-\csc^2x$
    ---------
    Thus given,
    $\displaystyle y=\ln| \csc x-\cot x|$
    Then by the chain rule you have,
    $\displaystyle y'=(\csc x-\cot x)'\cdot \frac{1}{\csc x-\cot x}$
    Thus,
    $\displaystyle y'=\frac{(-\csc x\cot x +\csc^2x)}{\csc x-\cot x}=\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}=\csc x$

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

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

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

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

    2. Find the antiderivative:
    $\displaystyle
    \int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt
    $

    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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