1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral sqrt((5-x)/x)

Tags:
  1. Dec 31, 2017 #1
    1. The problem statement, all variables and given/known data
    integration with respect to x

    2. Relevant equations

    integral 1/sqrt (a^2 - x^2) = arcsin(x/a)
    3. The attempt at a solution
    image attached, the arcsine term in 5/2 arcsin((2x-5)/5) it should be 5 arcsine(sqrt(x/5))
     
  2. jcsd
  3. Dec 31, 2017 #2
  4. Dec 31, 2017 #3

    Delta²

    User Avatar
    Homework Helper
    Gold Member

    I think the arcsine formula for computing ##\int \frac{1}{\sqrt{(\frac{5}{2})^2-(x-\frac{5}{2})^2}}dx## cannot be applied because for x=0 the denominator goes to 0.
     
    Last edited: Dec 31, 2017
  5. Dec 31, 2017 #4

    QuantumQuest

    User Avatar
    Science Advisor
    Gold Member

    It would be far better to use Latex in order to be helped because checking through the attached image is somewhat hard.
    Now, for the integral, I would hint in the way of substitutions with ##u = \frac{5 - x}{x}## being the most obvious. Have you tried this way?
     
  6. Dec 31, 2017 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your image is an unreadable mess. I cannot tell what you think the final answer should be. Please take the trouble to type out at least your final answer.
     
  7. Dec 31, 2017 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    When I plot those two functions in Mathematica, they appear to differ only by a constant, namely ##5\pi/4##.
     
  8. Dec 31, 2017 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You have the same issue for
    $$\int \frac{dx}{\sqrt{1-x^2} } = \arcsin x + C$$ at ##x=\pm 1##.
     
  9. Dec 31, 2017 #8

    Delta²

    User Avatar
    Homework Helper
    Gold Member

    Well for some reason I thought the domain of the function should contain 0, but now I understand this is not necessary. Should just state that the result holds for ##x\neq 0##
     
    Last edited: Dec 31, 2017
  10. Dec 31, 2017 #9
    Yeiks, they look so different though. looks like constants do indeed make a huuuge difference. Thank you though, this was giving me a headache.
     
  11. Dec 31, 2017 #10
    That substitution isn't working for me :/, I'm getting ## du =\frac{5}{x^2} dx## which gives me ## dx= \frac{1}{(u^2+1)^2} ## integral is this times a root u
     
  12. Jan 1, 2018 #11

    ehild

    User Avatar
    Homework Helper

    Try the substitution ##u=\sqrt{\frac{5-x}{x}}##
     
  13. Jan 1, 2018 #12
    This is u^2 =previous substitution . I'd get integral mentioned in previous reply but instead of multiplied by root u it'd be divided by 2 sqrt(u). any pointers on how i might use this more meaningfully?
     
  14. Jan 1, 2018 #13

    ehild

    User Avatar
    Homework Helper

    What is x if ##u^2=\frac{5-x}{x}## and what is dx? Is there any square root in the integrand?
     
  15. Jan 1, 2018 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you have an answer to an integral that you are not sure about, you can always differentiate it to see whether you get back the integrand.
     
  16. Jan 1, 2018 #15

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    If you have any textbook with a chapter or two on integration, you will find a list related things, practically a family of them, there.
    In this family it is quite common to have integrations that work only within certain range of x, a different formula outside it. This can correspond to something physical such as existence of escape velocity or runaway reaction outside the ‘tame’ region.

    Not everybody does it this way but I recommend to change the variable so as not to have the constant ##a## within the formula to be integrated, e.g let ##x = 5aX##, a new variable . That way you can get integrands that are more recognisable and reduced to a smaller standard set. (You cannot forget this original ##a##, especially when you also have to change the ##dx## and also at the end of the calculation .)

    There are then various ways, but I think what you were doing looks unnecessarily complicated than the simplest approach is just a further change of variable defining a ##y = sin X##

    As things like this do not usually come out of the blue, I wonder if this reminds you of anything in that lesson or book?
     
  17. Jan 1, 2018 #16

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    A general trick is to try to get rid of the square root by making a variable substitution. In the case of [itex]\sqrt{\frac{5-x}{x}}[/itex], if you let [itex]x = 5 cos^2(\theta)[/itex], then this becomes: [itex]\sqrt{\frac{1-cos^2(\theta)}{cos^2(\theta)}} = \frac{sin(\theta)}{\cos(\theta)}[/itex]
     
  18. Jan 1, 2018 #17

    QuantumQuest

    User Avatar
    Science Advisor
    Gold Member

    The substitution works but if you find it difficult to follow then try what ehild suggests in post #11. If you find ##dx## and substitute in the original integral for the new variable you'll get a manageable integral.
     
  19. Jan 2, 2018 #18
    integral is ## - \frac{2u^2du}{(u^2+1)^2} ## where u is what you suggested. This is to be followed by a substitution of u = tan z?
     
    Last edited: Jan 2, 2018
  20. Jan 2, 2018 #19

    ehild

    User Avatar
    Homework Helper

    I used integration by parts . ##f '= \frac{2u}{(u^2+1)^2} ##, g=u.
     
    Last edited: Jan 2, 2018
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted