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Math Help - calc 2 final

  1. #1
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    calc 2 final

    I am having trouble with a few areas in my calc 2 study. I do not completely understand the steps involved in my book, and I would like to fix this before my final tomorrow. (btw is latex working yet?)

    For example in the integral, int[ (4x^2+9)^(1/2) / x^4 ] dx

    I do not understand how to approach the problem at all. I know it is a trigonometric substitution, but when I look at my teachers notes I see she has the equation int[ (3*sec(t) ( 3/2 sec^2 (t)) dt) / ((3^4/2^4) (tan^4 (t))) ]dt as the next step.

    I have no idea where the (3/2 sec^2 (t)) came from. It looks like the first term (3*sec(t)) only integrated, but i know it hasn't been, because that isn't the integrated form of 3*sec(t). I also have no idea how she got her denominator.

    3^x/2^x * tan^x (t) ???

    Any help here or on trig subs in general would be nice





    Similarly on the trig integral: int[ (cos^4(x)) ], i follow it fine until my book states the coefficient of the following step:

    =1/4 int[dx] + 1/4 int[ (cos 2*x)(2)(dx) ] + 1/8 int [ (1+cos 4*x) dx ]

    where does the 1/8 come from here?



    the same problem is seen in the integral int [ sin^2(x) dx ]

    on the step = 1/2 int[dx] - 1/4 int [ cos 2x * 2 * dx ], i dont see where the 1/4 is coming from.



    thanks.
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  2. #2
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    Quote Originally Posted by thedoge View Post
    I am having trouble with a few areas in my calc 2 study. I do not completely understand the steps involved in my book, and I would like to fix this before my final tomorrow. (btw is latex working yet?)

    For example in the integral, int[ (4x^2+9)^(1/2) / x^4 ] dx
    Note, 4x^2 = (2x)^2

    So first use the substitution t=2x
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thedoge View Post

    Similarly on the trig integral: int[ (cos^4(x)) ], i follow it fine until my book states the coefficient of the following step:

    =1/4 int[dx] + 1/4 int[ (cos 2*x)(2)(dx) ] + 1/8 int [ (1+cos 4*x) dx ]

    where does the 1/8 come from here?
    Remember, when integrating sin^2(x) or cos^2(x) we change them to (1 - cos(2x))/2 and (1 + cos(2x))/2 respectively. when we factor out those halves, that's what causes the change in your constant multiplyers.

    int{cos^4(x)}dx = int{cos^2(x) * cos^2(x)}dx
    .......................= int{[1 + cos(2x)]/2 * [1 + cos(2x)]/2}dx
    .......................= (1/4)int{(1 + cos(2x))^2}dx
    .......................= (1/4)int{1 + 2cos(2x) + cos^2(2x)}dx
    .......................= (1/4)int{1}dx + (1/4)int{2cos(2x)}dx + (1/4)int{(1 + cos(4x))/2}dx
    .......................= (1/4)int{1}dx + (1/2)int{cos(2x)}dx + (1/8)int{1 + cos(4x)}dx
    can you take it from there?

    the same problem is seen in the integral int [ sin^2(x) dx ]

    on the step = 1/2 int[dx] - 1/4 int [ cos 2x * 2 * dx ], i dont see where the 1/4 is coming from.
    Similarly,
    int{sin^2(x)}dx = int{(1 - cos(2x))/2}dx
    ......................= (1/2)int{1 - cos(2x)}dx
    ......................= (1/2)int{1}dx - (1/2)int{cos(2x)}dx

    now for some reason, the book saw it fit to multiply by 2/2, it split it into (1/2)*2 and left the (1/2) outside, which gave you the 1/4, then brought the 2 inside, which gave you the cos(2x) * 2. then i suppose they wanted to make the subtitution u = 2x, and then they could use the 2 dx that way. can you take it from here?

    This is my 15 th post!!! Yet i'm still "just really nice"
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    aww, were you hoping for more than just being nice?

    hrm well yea that helps, but i have a few more questions

    How would I reduce sin(sec^-1 x) <-- not a Dx or integral

    How would I find the derivative of (sin x)^(tan x) <-- i assume natural log of both sides

    How would I find the derivative of ln(3x/(x-3)) <---i assume i take 1/u*du after taking ln(3x)-ln(x-3) <-- ln(a/b) = ln a -ln b is a ln property, i think

    How would I take the limit of (1/e^x-1)-(1/x) <-- i know I need to take l'hopital, but i am getting tricked up with the 2nd l'hopital... i'd like to see it done properly

    thanks again
    Last edited by thedoge; May 2nd 2007 at 01:45 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thedoge View Post
    aww, were you hoping for more than just being nice?
    of course i want to be more than nice. i think i should be a "glorious beacon of light" by now. dont know what's the hold up. maybe i don't have enough posts

    How would I reduce sin(sec^-1 x) <-- not a Dx or integral
    let sec^-1x = theta
    => x = sec(theta)
    so draw a right angle triangle. with an acute angle called theta, the hypotenuse = x and the adjacent side to theta = 1, then the other side (by pythagoras) = sqrt(x^2 - 1)

    so we have, sin(sec^-1x) = sin(theta) = opp/hyp = sqrt(x^2 - 1)/x

    is that what you wanted?

    How would I find the derivative of (sin x)^(tan x) <-- i assume natural log of both sides
    yup, that's how i'd do it.

    y = (sinx)^(tanx)
    => lny = tanx * ln(sinx)
    => y'/y = (sec^2x)ln(sinx) + (tanx)*(1/sinx)*cosx = (sec^2x)ln(sinx) + 1
    => y' = [(sec^2x)ln(sinx) + 1]*y
    => y' = [(sec^2x)ln(sinx) + 1]*(sinx)^(tanx)

    and you can simplify if you wish

    How would I find the derivative of ln(3x/(x-3)) <---i assume i take 1/u*du and use quotient rule?
    you could do that, but i'd use the law of logs to simplify first, that way, the quotient rule isn't needed.

    y = ln(3x/(x - 3))
    => y = ln(3x) - ln(x - 3)
    => y' = (1/x) - 1/(x - 3)

    i just noticed you mentioned a substitution 1/u * du, did you mean to say take the integral of, becuase u-substitution is not a differentiation technique

    How would I take the limit of (1/e^x-1)-(1/x) <-- i know I need to take l'hopital, but i am getting tricked up with the 2nd l'hopital... i'd like to see it done properly
    the limit as x goes to what? zero? infinity?

    and is it 1/[(e^x) - 1] or 1/(e^(x - 1))?

    well, in either case, l'hopital's doesn't seem to be necessary
    Last edited by Jhevon; May 2nd 2007 at 01:37 PM.
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  6. #6
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    the last one is the lim as x-> 0 and [1/(exp(x)-1)] - [1/x]

    how isn't l'hopital necessary? heh

    i was taught that when you are taking a limit of two separate differentiable functions that you cross multiply them to get a new quotient, then apply l'hopital if necessary.

    so (1*x)-(1*exp(x)-1) / (x*(exp(x)-1)) = [x+1-exp(x)] / [xexp(x)-x] which evalulated @ 0 give you 0/0.

    how else would tis be done?

    and yeah, i noticed on the second to last one i made an error.


    On the first one, i'm not 100% sure if thats what I need or not. It is a problem dealing with the sin of the (arcsecant (x)). I got it wrong on the test (first test, first problem-woo) and never figured out the right answer.
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    Quote Originally Posted by Jhevon View Post
    i just noticed you mentioned a substitution 1/u * du, did you mean to say take the integral of, becuase u-substitution is not a differentiation technique

    Nope. I meant Dx of Ln(u) = du/u . For a bit you had me thinking I was wrong, but I looked it up in my book again to confirm this

    to prove it, look at Dx Ln(3x) = 1/x. But why? 1/3x*3 = 3/3x= 1/x.

    u=3x
    1/u=1/3x
    du=3
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