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!

Integrating dx / (4+x^2)^2 using Trigonometric Substitution

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral:
    integral of dx / (4+x^2)^2

    2. Relevant equations
    x = a tan x theta
    a^2 + x^2 = a^2 sec^2 theta

    3. The attempt at a solution
    x = 2 tan theta
    dx = 2sec^2 theta
    tan theta = x/2

    integral of dx / (4+x^2)^2
    = 1/8 integral (sec^2 theta / sec^4 theta) d theta
    = 1/8 integral (cos^2 theta) dtheta
    = 1/16 integral 1 + cos 2 theta d theta
    = 1/16 [ theta + sin 2 theta / 2 ] + c
    = 1/16 [ theta + sin theta cos theta ] + c
    = 1/16 [ tan ^-1 x/2 + ??? ] + c

    "???" is where I get lost.
    Did I do the problem wrong? I was really confused when the book gave me a problem without a square root.
     
  2. jcsd
  3. Oct 25, 2015 #2

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Your sub looks fine, and your work up to $$=\frac{\theta}{16}+\frac{sin(2\theta)}{32}+C$$

    Go ahead and make your last back sub.
     
  4. Oct 25, 2015 #3
    a^2 + x^2 = a^2 sec^2 theta
    x^2 + 4 = 4 sec^2 theta
    sqrt (x^2 + 4) / 2 = sec theta
    cos sqrt (x^2 + 4) / 2 = theta

    Is this what you mean by back sub?
     
  5. Oct 25, 2015 #4

    Student100

    User Avatar
    Education Advisor
    Gold Member

    You have ##\theta = tan^{-1}(\frac{x}{2})## above, just plug in for theta and simplify.
     
  6. Oct 25, 2015 #5
    So,

    = [ θ/16 + sin (2θ)/32 ] + C
    = [ tan^-1 (x/2) / 16 + sin (2 tan^-1 (x/2) / 32 ] + C

    Is it okay to plug in tan^-1 (x/2) for θ in sin (2θ) or am I misinterpreting you?
    It seems weird.
     
  7. Oct 25, 2015 #6

    Student100

    User Avatar
    Education Advisor
    Gold Member

    It's perfectly fine, that's what you needed to do. You had already found above that ##tan(\theta)=\frac{x}{2}##, you could simplify the sin by redoing the double angle formula ##sin(2tan^{-1}(\frac{x}{2}))=2(sin(tan^{-1}(\frac{x}{2}))(cos(tan^{-1}(\frac{x}{2}))## and simplify from your right triangle picture.

    Why does it seem weird?
     
  8. Oct 25, 2015 #7
    = [ tan^-1 (x/2) / 16 + (sin(4/x) cos(4/x))/16 ]

    Does it simplify to that?

    It seems weird because I don't think I've done it before and the answer in the back of the book is in a different format:
    (tan^-1 (x/2) )/ 16 + x/(8(4 + x^2)

    Sorry for the messy and inconsistent typing, and thanks for the help.
     
  9. Oct 25, 2015 #8

    Student100

    User Avatar
    Education Advisor
    Gold Member

    No problem, so you havehttps://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/73/73867-db7eccc9bcdd5e7324f0dfba67853800.jpg [Broken]
    $$ = \frac{1}{16}(tan^{-1}(\frac{x}{2})+(sin(tan^{-1}(\frac{x}{2}))(cos(tan^{-1}(\frac{x}{2})))$$

    Since $$ \theta=tan^{-1}(\frac{x}{2})$$
    $$cos(tan^{-1}(\frac{x}{2}))=\frac{2}{\sqrt{x^2+4)}}$$

    Do the same for sin, simplify, and you should get the book answer.
     
    Last edited by a moderator: May 7, 2017
  10. Oct 25, 2015 #9
    I did it! Looks exactly like the book's.

    Thanks again.
     
  11. Oct 25, 2015 #10

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Anytime, you did all the calculations and setting up correctly, I think the end just through you off.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted