My Math Forum Sqrt of (tan x)^2

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 January 7th, 2019, 11:35 PM #1 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Sqrt of (tan x)^2 When solving the integral of sqrt(x^2-a^2) where a is a positive integer and letting x = a*sec theta, there is a step of doing the square root of (tan theta)^2. For pi/2 < theta < pi, the square root would be -tan theta. My question is why there is a minus sign before the trigonometric expression if tan theta is already a negative value for pi/2 < theta < pi? Moreover, if the range of theta is not specified, is the square root equal to tan theta only? And why? Last edited by skipjack; January 8th, 2019 at 02:56 AM.
 January 8th, 2019, 02:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 The expression sqrt(x² - a²), where x² - a² > 0, means the positive square root of (x² - a²). When a substitution is made, the domain for the new variable needs to be chosen (explicitly or implicitly). If, for your example, that domain is pi/2 < θ < pi, the expression sqrt(x² - a²) becomes sqrt(a²tan²θ), which, as a > 0 and tan θ < 0, simplifies to -a*tan θ. Thanks from justusphung and topsquark
 January 8th, 2019, 05:54 PM #3 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 I would like to make sure of some basic concept. If the domain of the substitution is not specified, like (x^2)^(3/2) which is simply x^3, but when x<0, then the answer should be (-x)^3, right? Therefore, when the domain of x is not clearly said, how can I conclude that (x^2)^(3/2) is x^3, but not (-x)^3
 January 8th, 2019, 10:19 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 For x < 0, (x²)^(3/2) = -x³, not x³. If you wish to find an indefinite integral of some function of y with respect to y, and make the substitution y = f(x), the function f(x) must be invertible and clearly defined. However, if, say, the substitution is y = x³, it's well-known that this function of x is clearly defined and invertible for all real x, so it would be acceptable to leave the domain of f(x) unstated, but implicitly "all real x". That wouldn't apply to the substitution y = sec(x), as sec(x) isn't invertible unless an appropriately restricted domain is chosen. Similarly, (x²)^(3/2) isn't invertible unless an appropriately restricted domain is chosen. Particular care should be taken if you are integrating with respect to a variable that can't be assumed to be real. When substituting y = f(x), one may occasionally need to allow x to be imaginary, even though y is real, so as to ensure that f(x) is invertible.
January 8th, 2019, 10:57 PM   #5
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Thank you for your detailed explanations. So, it is important to indicate the domain of an equation. What about the steps shown in the pic? It assumes that x=sec theta/2 without stating the range of theta. And it simplifies the term of ((sec theta)^2 - 1)^(3/2) into (tan theta)^3 with no consideration about the negative values of tan theta. Is it more acceptable to write abs(tan theta)^3 instead?
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 January 9th, 2019, 05:46 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 If the domain is chosen as pi/2 < θ < pi, so that sec θ and tan θ are negative, (sec²θ - 1)^(3/2) = -tan³θ. The integral becomes (1/2)cosec θ + C instead of -(1/2)cosec θ + C. One now has (1/2)cosec θ = -x/√(4x² - 1), not x/√(4x² - 1), so the final answer is the same as in your pic. However, this doesn't justify the presentation in that pic. The author probably had in mind the domain 0 < θ < pi/2, but should have given the domain explicitly. Thanks from justusphung
 January 9th, 2019, 08:36 PM #7 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 When there is an arcsec in the primitive function, the book does state the range of theta, but not in this case. The author should clearly indicate the domain of x to avoid ambiguity as you have mentioned. Anyway, thanks for your help and patience.

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