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Math Help - Trigonometric substitution problem

  1. #1
    s3a
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    Trigonometric substitution problem

    The initial problem is: http://www.wolframalpha.com/input/?i=integrate(sqrt(x^2+-+1)%2Fx%2C+x%2C1%2C2)

    My problem is not in the substitution part but rather on the limits of integration. I tried "an easy way out" which is to just use => instead of = signs and not put limits of integration and then just turn the variables into x variables (initial variables) and use the initial limits of integration but the problem I am having is from this step:

    integral of (tan^2 (theta) (dtheta)) = tan(theta) - (theta) + C = sqrt(x^2 - 1) - arcsec(x) and when I try to find the value for the final answer:

    sqrt(2^2 - 1) - arcsec(2) - 0 + arcsec(1)

    both bold parts give me errors on the calculator which means they do not exist. I do recall that theta should be between [0, pi/2) or [pi, 3pi/2) but there is a final answer in the back of the book (sqrt(3) - pi/3)which suggests I am doing something wrong.

    Any help would be GREATLY appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    The initial problem is: http://www.wolframalpha.com/input/?i=integrate(sqrt(x^2+-+1)%2Fx%2C+x%2C1%2C2)

    My problem is not in the substitution part but rather on the limits of integration. I tried "an easy way out" which is to just use => instead of = signs and not put limits of integration and then just turn the variables into x variables (initial variables) and use the initial limits of integration but the problem I am having is from this step:

    integral of (tan^2 (theta) (dtheta)) = tan(theta) - (theta) + C = sqrt(x^2 - 1) - arcsec(x) and when I try to find the value for the final answer:

    sqrt(2^2 - 1) - arcsec(2) - 0 + arcsec(1)

    both bold parts give me errors on the calculator which means they do not exist. I do recall that theta should be between [0, pi/2) or [pi, 3pi/2) but there is a final answer in the back of the book (sqrt(3) - pi/3)which suggests I am doing something wrong.

    Any help would be GREATLY appreciated!
    Thanks in advance!
    Are you sure you took "arcsec" rather than "arcsin" or "arccos"? The sine and cosine values are always between -1 and 1 so arcsin(2) or arccos(2) do not exist (though I can't imagine why "arcsin(1)" or "arccos(1)" would be a problem) but since secant is "1 over cosine" arcsec(2) is the same as arccos(1/2) which is \pi/3 and, of course, arcsec(1) is the same as arccos(1) which is 0.
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  3. #3
    s3a
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    I had tried arcsec(2) = 1/arccos(2) and arcsec(1) = 1/arccos(1). So, what I have to do instead is take the reciprocal of the fraction and then use the reciprocal trig?

    By this I mean, on the first example, arcsec(2) = arcsec(2/1) = arccos(1/2) ?

    and for the second;

    arcsec(1/1) = arccos(1/1) ?
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    Quote Originally Posted by s3a View Post
    I had tried arcsec(2) = 1/arccos(2) and arcsec(1) = 1/arccos(1). So, what I have to do instead is take the reciprocal of the fraction and then use the reciprocal trig?

    By this I mean, on the first example, arcsec(2) = arcsec(2/1) = arccos(1/2) ?

    and for the second;

    arcsec(1/1) = arccos(1/1) ?
    I find it helpful to think of arcsec(2) as the angle whose secant is 2. So if the secant of the angle is 2 then the cosine of the angle is 1/2 and

    arcsec(2) = arccos(1/2)
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