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Can this be solved?

  1. Feb 1, 2005 #1
    I was given this problem and the persom whom I recieved it from told me they couldn't see how it would he solved or even if it had an answer. I think I might have come to a conclusion even thought it might be wrong, what do you guys think

    [tex]\int tan^2x sec^3x dx[/tex]

    this is what i've been able to conclude, might be wrong but I gave it a shot

    [tex]\int tan^2x sec^3x = \int (sin^2x/cos^2x)(1/cos^3x) dx[/tex]
    [tex]\int (sin^2x)(cos x)/(cos^3x) dx = \int sin^2x/cos^2x = \int tan^2x[/tex]
    [tex]= sec^2x + c[/tex]

    Im guessing thats probably wrong, but what do you guys think?
  2. jcsd
  3. Feb 1, 2005 #2
    You made a mistake when you "cancelled" the cosines.

    [tex]\frac{\sin^2{x}}{\cos^2{x}} \cdot \frac{1}{\cos^3{x}} = \frac{\sin^2{x}}{\cos^5{x}}[/tex].
  4. Feb 1, 2005 #3
    oops for some reason my brain was thinking common denominator, ill try to see if I can solve it with what you just gave.. I figured if their was a mistake it was at that step.

    edit: BTW can anyone else tell me if Muzza is right and I was wrong at that step? im thinking its correct but like I said some of you guys know a lot more then me.
    Last edited: Feb 1, 2005
  5. Feb 1, 2005 #4
    yes it can be solved

    Muzza is correct...

    Last edited: Feb 1, 2005
  6. Feb 1, 2005 #5


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    Since it is not in the HW section,i'll solve most of it...

    [tex] I=\int \tan^{2}x \sec^{3}x dx=\int \frac{\sin^{2}x}{\cos^{5}x} dx [/tex] (1)

    Use part integration to get:

    [tex] I=\sin x \cdot (-1)\frac{\cos^{-4}x}{-4}-\frac{1}{4} \int \frac{dx}{\cos^{3}x} dx [/tex] (2)

    Denote the last integral by J and it is computed as follows:

    [tex] J=\int \frac{\cos x}{\cos^{4}x} dx=\int \frac{d(\sin x)}{(1-\sin^{2}x)^{2}} [/tex] (3)

    Making the obvious substitution
    [tex] \sin x =u [/tex] (4)
    ,"J" becomes:
    [tex]J=\int \frac{du}{(1-u^{2})^{2}} =\int \frac{du}{(1+u)^{2}(1-u)^{2}} [/tex] (5)

    Hopefully the decomposition into simple fractions that i made is correct:

    [tex] \frac{1}{(1+u)^{2}(1-u)^{2}}=\frac{1}{4}\frac{u}{(1+u)^{2}}+\frac{1}{2}\frac{1}{(1+u)^{2}}-\frac{1}{4}\frac{u}{(1-u)^{2}}+\frac{1}{2}\frac{1}{(1-u)^{2}} [/tex] (6)

    I'll let you compute the 4 elementary integrals that would yield "J" in terms of "u" and then invert the substitution (4) and finally compute "I" from (2).

    Good luck!!

    Last edited: Feb 1, 2005
  7. Feb 1, 2005 #6


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    Dearly Missed

    An alternative to Daniel's approach, is to use the [tex]u=tan(\frac{x}{2})[/tex] substitution.
    Since you chose not to post in the homework section, I won't give you the solution..:wink:
  8. Feb 1, 2005 #7


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    Guess what,Arildno...Wolfram Integrator's Mathematica 5.0 reccomends the same "tangent of half-angle" substitution... :tongue2:

    Did u have a contribution at that software's birth...?? :confused: :wink:


    P.S.I knew about the solution Mathematica offered before "cooking" the solution... :cool: But i found it :yuck:
  9. Feb 1, 2005 #8
    Hey dexter im sorry but you sort of lost me at step 2 when you did the integration by parts. Were you using this formula(this is off the top of my head might be slightly different):

    [tex]\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx ?[/tex]

    Also im not 100% sure how to implement this at the moment but if the top if the direvative of the bottom cant you implement the natural log (1/u)? im not sure if that can be done with this case.

  10. Feb 1, 2005 #9


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    Yes,of course...

    Well,at the moment i used,no...But in the progress of computing the 4 simple integrals,yes,natural logarithm pops up...

  11. Feb 1, 2005 #10
    Couldn't the integral be solved more easily with substitution?

    [itex]\int (\sec^2x - 1)\sec^3x dx[/itex]

    [itex]\int \sec^5x - \sec^3x dx[/itex]

    and then applying power reduction formulas... I'm not sure, because I don't know what they are off-hand, but I remember evaluating integrals similar to that by using subbing a simple trig identity as the first step.
    Last edited: Feb 1, 2005
  12. Feb 1, 2005 #11


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    I don't know...Why don't u solve it and compare to my solution??And then with Mathematica's solution (half angle substitution)...

  13. Feb 1, 2005 #12
    that to me looks like it can also work, im feeling really lazy right now or i'd solve it, ill try it tomorrow.

    Thanks for the assistance guys.
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