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Integral Calculus

  1. Aug 26, 2008 #1
    It has been a while since I have had to do an integration such as this, it is probably quite simple lol. But could someone show me how to integrate (secx)^2, that is secx all squared. I know that the answer is tanx but i was wondering if someone could show me the method.

    Thank you.
     
  2. jcsd
  3. Aug 26, 2008 #2

    Defennder

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    Easiest way to do so is to differentiate tan x = sin x/ cos x by quotient rule, thus avoiding all the integration.
     
  4. Aug 26, 2008 #3
    That is a valid point thank you, however, I am specifically looking for the integration solution.
     
  5. Aug 26, 2008 #4
    Hmm, well I haven't taken many college mathematics classes yet, but I suspect that by taking an analysis class one could prove why [tex]\int[/tex]sec2(x) dx=tan(x)+c. I think of it like this: I know (and can show) that the derivative of sin(x) is cos(x). Therefore, the antiderivative of cos(x) is sin(x)+c. Likewise, since I know that the derivative of tan(x) is sec2(x), the integral of sec2(x) must be tan(x)+c.
     
  6. Aug 26, 2008 #5
    Well it's easy if we're allowed to use the fact that the derivative of tan(x) is [tex]sec^2(x)[/tex]. Just try a u-sub or use the pythagorean identity and then u-sub. But otherwise, I don't have any clever ideas.
     
  7. Aug 26, 2008 #6
    The OP isn't questioning that you could "prove" it by taking derivative of tan(x) + C. He just wants to know how to do the integral if we did not know what [tex] sec^{2}(x) [/tex] integrates to.
     
  8. Aug 26, 2008 #7
    Ok the way i know is like this.
    sec^2(x) = tan(x) then sec^2(x) dx = tan(x) + c
     
  9. Aug 26, 2008 #8
    that made no sense.
     
  10. Aug 26, 2008 #9
    Errr what? [tex]sec^2(x) = tan^2(x) + 1[/tex]

    If you substitute u = sec(x) and then perform a subsequent trig substitution, you could find the integral provided that you're okay with using the fact that the derivative of [tex]tan(x)[/tex] is [tex]sec^2(x)[/tex].

    I think you would be hard-pressed to find a solution that doesn't use the fact that the derivative of [tex]tan(x)[/tex] is [tex]sec^2(x)[/tex] and though you might gain something from integration via substitution, it's kind of overkill since most people usually learn derivatives of trig functions first.
     
  11. Aug 26, 2008 #10
    Hmmm, let's see. Let's use the shorthand notation:

    s = sin(x)

    c = cos(x)

    We want to integrate 1/c^2, in a "canonical way", i.e. using the usual rules of manipulation we always use and not using any ad hoc rules that only apply in this case, like recognizing that the answer is tan(x) and proving that by differentiation.

    We can simply proceed in a similar way as how we would derive the reduction formulas for integration of 1/c^n, c^n, etc. etc:

    1/c^2 = (c^2 + s^2)/c^2 = 1 + s^2/c^2

    We now integrate s^2/c^2 using partial integration:

    We integrate the factor s/c^2. This yields:

    s/c - Integral of [1/c times derivative of s] dx = s/c - x

    So, the integral of 1/c^2 is s/c

    Note that if you replace 1/c^2 by 1/c^n, the above method yields a relation between the integral of 1/c^n and 1/c^(n-2).
     
  12. Aug 26, 2008 #11
    I know, I wasn't trying to offer a proof though. I was just demonstrating my reasoning. Am I correct in saying that his question could be answered in some sort of real analysis class though?
     
  13. Aug 27, 2008 #12

    HallsofIvy

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    It's not clear to me why you post here if you refuse to take the advice given. Defennder told you exactly how to find the integral.
     
  14. Aug 28, 2008 #13
    Here's what I tried but I cannot find my mistake:

    [tex] \displaystyle{\int}\frac{1}{\cos^2x} dx=\displaystyle{\int}\frac{1}{(Re\{e^{ix}\})^2} dx=Re \{\displaystyle{\int}\frac{1}{e^{2ix}} dx\}=Re\{\displaystyle{\int}e^{-2ix} dx \}=Re\{-\frac{1}{2i}e^{-2ix}\}[/tex]

    [tex]Re\{-\frac{1}{2i}e^{-2ix}\}=Re\{\frac{2i}{4}[\cos(-2x)+i\sin(-2x)]\}=Re\{\frac{2i}{4}[\cos(2x)-i\sin(2x)]\}=\frac{\sin2x}{2}=\sin x\cos x+C[/tex]

    My result is most obviously incorrect. Can anyone help me find the error? Thanks
     
  15. Aug 28, 2008 #14
    You cannot take the Re operator out of the integral sign from the denominator.
     
  16. Aug 28, 2008 #15
    Ok, I see, but if it were in the numerator? I suppose it would be possible then, wouldn't it?
     
  17. Aug 28, 2008 #16
    In general, it only works if the Re sign is in front of the entire expression. If the denominator is a real function, then you can pull a Re sign from the numerator and place it in front of the expression.

    You should try to prove this (e.g. using that the integral of a sum of two terms is the sum of the integrals of the two terms)
     
  18. Aug 28, 2008 #17

    HallsofIvy

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    Whatever happened to Q.E.D.? The first response didn't do the problem for him so he went off to sulk?
     
  19. Aug 29, 2008 #18

    Gib Z

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    Let [itex]t=\tan \left( \frac{x}{2} \right)[/itex], after simplifying the original integral directly to [tex] 2\int \frac{1+t^2}{ (1-t^2)^2 } dt[/tex] use partial fractions and its quite straight forward from there.
     
  20. Aug 29, 2008 #19

    Defennder

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    There's one problem here: You need to be able to replace dx with dt. But to do so you need differentiate arctan. How would you differentiate arctan without using d/dx tan x = sec^2 x ?
     
  21. Aug 29, 2008 #20
    Indeed. So, I gave the only correct solution earlier on in this thread: use the same reduction formula you use when you want to integrate 1/cos^n(x).
     
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