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!

Factorizing denominator help

  1. Jun 21, 2005 #1
    [tex]\int\frac{1}{(\sin(x)+\frac{2}{\cos(x)})^2} dx[/tex]
    I know that tan(x)=t works, but there are some problems with factorizing denominator and even after factorizing that integral isn't very simple. Does anyone see here any simplier substitution? thank you
     
  2. jcsd
  3. Jun 21, 2005 #2
    Its [itex]tan(\frac{x}{2}) = t[/itex]. Try it. What are the problems?
     
  4. Jun 21, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Don't be afraid of this simple algebra.When doing integrals,you ought to be prepared for worst.

    BTW,LaTex acknowledges the function "tangent",as long as you write it " \tan ".

    Daniel.
     
  5. Jun 21, 2005 #4
    In this case denominator is [tex]t^4-t^3+2t^2+t+1[/tex]. This i can't factorize also. Even Mathematica doesn't help.
     
  6. Jun 21, 2005 #5
    Hm. Yes, i'm probably just too lazy today to work on algebra. But i have one more question about such problems. If i have polynomials with complex solutions what is the algorithm for factorizing them?
    As examples:1) [tex]t^4-t^3+2t^2+t+1[/tex]
    2) [tex]4t^4+4t^3+9t^2+4t+4[/tex]
     
  7. Jun 21, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It can't be.

    [tex]I= \int \frac{dx}{\left(\sin x+\frac{2}{\cos x}\right)^{2}} [/tex] (1)

    Make the sub

    [tex] \tan\frac{x}{2}=t [/tex] (2)

    An you'll get the integral

    [tex] I=2\int \frac{\cos ^2\left( 2\arctan t\right) }{\left[ \sin \left( 2\arctan t\right) \cos \left( 2\arctan t\right) +2\right] ^2\left( 1+t^2\right) }\,dt [/tex] (3)

    Then

    [tex] \left\{\begin{array}{c} \cos \left( \arctan t\right) =\frac{1}{\sqrt{\left( 1+t^2\right) }}\\ \sin \left( \arctan t\right) =\frac {t}{\sqrt{\left( 1+t^2\right) }} \end{array} \right [/tex] (4)

    Can you take it from here...?

    Daniel.
     
    Last edited: Jun 21, 2005
  8. Jun 21, 2005 #7
    Yeah I forgot. Thanks anyway.
     
  9. Jun 21, 2005 #8
    [tex] \left{\begin{array}{c} \cos \left( 2\arctan t\right) =\frac{1-t^2}{{\left( 1+t^2\right) }}\\ \sin \left( 2\arctan t\right) =\frac {2t}{{\left( 1+t^2\right) }} \end{array} \right [/tex]
     
  10. Jun 21, 2005 #9
    Yes, i did it dextercioby. This i have from it:
    [tex]\int\frac{(1+t^2)(1-t^2)^2}{t^4-t^3+2t^2+t+1}[/tex]
    OK. i can devide this fraction, but i can do nothing with denominator
     
  11. Jun 21, 2005 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Here's what i'd do.I denote your integral by "I".

    [tex] I=:\int\frac{\cos^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx [/tex] (1)

    I define its sister integral

    [tex] J=:\int\frac{\sin^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx [/tex] (2)

    The difference

    [tex] I-J=\int \frac{\cos 2x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx =\int \frac{d\left(\frac{1}{2}\sin 2x +2\right)}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} =-\frac{1}{\frac{1}{2}\sin 2x +2} [/tex] (3)

    I'll concentrate on the sum.

    To be continued.

    Daniel.
     
  12. Jun 21, 2005 #11

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Here's the sum

    [tex] I+J=\int \frac{dx}{\left(\frac{1}{2}\sin 2t +2\right)^2} [/tex] (4)

    I make the sub

    [tex] 2x=t [/tex] (5)

    [tex] I+J=\frac{1}{2}\int \frac{dt}{\left(\frac{1}{2}\sin t +2\right)^{2}} [/tex] (6)

    I make the sub

    [tex] \tan\frac{t}{2}=u [/tex] (7)

    [tex] I+J=\int \frac{1+u^{2}}{(2u^{2}+u+2)^{2}} \ du=\frac{1}{4}\int \frac{1+u^{2}}{\left[\left(u+\frac{1}{4}-\frac{i\sqrt{3}}{2}\right)\left(u+\frac{1}{4}+\frac{i\sqrt{3}}{2}\right)\right]^{2}} \ du [/tex] (8)

    Therefore

    [tex] I+J=\int \frac{1+u^{2}}{\left[u^{2}+u\left(\frac{1}{2}-i\sqrt{3}\right)-\frac{11}{16}-\frac{i\sqrt{3}}{4}\right]\left[u^{2}+u\left(\frac{1}{2}+i\sqrt{3}\right)-\frac{11}{16}+\frac{i\sqrt{3}}{4}\right]} \ du [/tex] (9)

    and now apply simple fractions.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Factorizing denominator help
Loading...