1. Not finding help here? Sign up for a free 30min 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!

Indefinite Integral - I

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}} \frac{\sec x}{\sqrt{1+2\sec x}}dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    The integral can be simplified to:
    [tex]\int \sqrt{\frac{1-\cos x}{1+\cos x}} \frac{1}{\sqrt{\cos x} \sqrt{\cos x+2}}dx[/tex]
    Using ##\cos x=2\cos^2x/2-1=1-2\sin^2x/2##, I end up with
    [tex]\int \frac{\tan(x/2)}{\sqrt{2\cos^2(x/2)-1}\sqrt{2\cos^2(x/2)+1}}dx=\int \frac{\tan(x/2)}{\sqrt{4\cos^4(x/2)-1}}dx[/tex]
    I am stuck here.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Apr 28, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Your algebra has a slight error at the beginning of item 3. above. sec x = 1 / cos x. When this is combined with 1 / SQRT (cos x + 2), you should obtain 1 / SQRT (cos^2 x * (cos x + 2)).
     
  4. Apr 28, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't see an error there.
    Pranav-Arora, you can eliminate the trig, if it helps. tan(x)dx = -d(cos(x))/cos(x). (I think that after a few substitutions you can get it to integrating sech.)
     
  5. Apr 28, 2013 #4
    Substituting x/2=t, dx=2dt
    [tex]2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt[/tex]
    ##\because \tan t=-d(\cos x)/(\cos x)##

    [tex]2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt=\int \frac{-d(\cos x)}{\cos x\sqrt{\cos^4(t)-(1/2)^2}}[/tex]

    Is this equivalent to integrating ##\displaystyle \int \frac{-dt}{t\sqrt{t^4-a^2}}## where a is some constant.

    I am asking this because I have never used this type of method to solve integrals.
     
  6. Apr 28, 2013 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    sec x is still not equal to 1 / SQRT(cos x).
     
  7. Apr 28, 2013 #6
    I never wrote that it is.
    I should have not skipped the steps. I simplified it the following way:
    [tex]\frac{\sec x}{\sqrt{1+2\sec x}}=\frac{1}{\cos x \sqrt{1+2/\cos x}}=\frac{1}{\sqrt{\cos x} \sqrt{2+\cos x}}[/tex]
     
  8. Apr 28, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Good. Now, what can you do after rewriting that as ##\int \frac{-tdt}{t^2\sqrt{t^4-a^2}}##?
     
  9. Apr 29, 2013 #8
    Let t^2=z or 2tdt=dz. The integral can be written as
    [tex]\frac{-1}{2} \int \frac{dz}{z\sqrt{z^2-a^2}}[/tex]
    My notes say that the above integral evaluates to -1/(2a)(arcsec(x/a)) but the answer is in terms of arcsin. :confused:
     
  10. Apr 29, 2013 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    well, arcsec(x/a) = arccos(a/x), and there's a pretty easy step from there to arcsin form.
     
  11. Apr 30, 2013 #10
    Furthermore, if you draw the right angled triangle and label the x and a sides correctly, you can relate all six trig functions to each other in terms of x and a.
     
  12. Apr 30, 2013 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I know that we're past this but there is a significant error in the middle of this post.

    You wrote that ##\ \tan t=-d(\cos x)/(\cos x)\ .##

    That should have been ##\ \tan(t)\,dt=-d(\cos t)/(\cos t)\,,\ ## which makes the next line:

    [itex]\displaystyle 2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt=\int \frac{-d(\cos t)}{\cos (t)\ \sqrt{\cos^4(t)-(1/2)^2}}\ .[/itex]​

    .
     
  13. May 1, 2013 #12
    Ah, I should have taken care of that. I did not notice that while I was writing that post. Thank you!

    Substituting cos(t)=z

    [tex]\int \frac{-d(\cos t)}{\cos (t)\ \sqrt{\cos^4(t)-(1/2)^2}}=\frac{-1}{2} \int \frac{dz}{z\sqrt{z^2-(1/2)^2}}=-\sec^{-1}(2z)+C=-\arccos\left(\frac{1}{2z}\right)+C[/tex]
    Adding and subtracting ##\pi/2##
    [tex]\frac{\pi}{2}-\arccos\left(\frac{1}{2z}\right)-\frac{\pi}{2}+C[/tex]
    [tex]\arcsin\left(\frac{1}{2z}\right)+K=\arcsin\left(\frac{1}{2}\sec^2\frac{x}{2}\right)+K[/tex]

    Thank you haruspex and SammyS! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Indefinite Integral - I
  1. Indefinite Integral (Replies: 1)

  2. Indefinite integration (Replies: 1)

  3. Indefinite Integration (Replies: 11)

  4. Indefinite Integrals (Replies: 4)

  5. Indefinite Integral (Replies: 2)

Loading...