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  1. Jan 1, 2007 #1

    Gib Z

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    Hey guys I was just wondering If you guys could post over a few indefinite integrals for me to solve, because I seem to find some of the integrals on these forums easy, and others im completely lost, I just want to see where im up to so I can start learning from there, my learning of calculus has been independant, so it hasnt been structured very well...

    Just post maybe a few at a time, with varying difficulty and I'll try to do them. I would prefer to get advice and do them on this thread rather than learn from a link if you guys dont mind...
     
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  3. Jan 1, 2007 #2

    ssd

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    Start with one of 12th standard level (it was in our text): sqrt(tanx)
     
    Last edited: Jan 1, 2007
  4. Jan 1, 2007 #3

    Hootenanny

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    Slightly more difficult one; Find I if

    [tex]I = \int \frac{dx}{2+\sin(x)}[/tex]
     
  5. Jan 1, 2007 #4

    Gib Z

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    O my jesus...I am completely lost on both of those, but from what I see on the mathematicia integrator, which doesnt show me steps btw..The solution to ssd's is way out of my leauge, and i havent even bothered to look up Hootenannys if its harder..Could you gives give me a prod in the right direction? Usually when I answer peoples integrals on this site I've read the hints others have given before...Im thinking maybe t=tan(x/2) substitution?
     
  6. Jan 1, 2007 #5

    cristo

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    Why not try the substitution u2=tanx ?

    Hint: Remember that sec2x=1+tan2x
     
    Last edited: Jan 1, 2007
  7. Jan 2, 2007 #6
    I would start with a change of bounds to simplify the integral. That is to say, find the integral of cosec(x).
     
  8. Jan 2, 2007 #7
    How exactly do you plan on doing that?
     
  9. Jan 2, 2007 #8
    Dah! Forget it. For some reason I thought it was 1/sin(x+2). My bad.
     
  10. Jan 2, 2007 #9
    Here something close:
    intergrat (1/(1+sin(u)) du = tan u - sec u +C

    I'm having trouble with LaTeX code reference loading up.
     
  11. Jan 2, 2007 #10
    Did you remember to put your code between [tex]\text{[tex]}[/tex] and [tex]\text{[\//tex]}[/tex] tags?
     
    Last edited: Jan 2, 2007
  12. Jan 2, 2007 #11
  13. Jan 2, 2007 #12

    Hootenanny

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    My apologies Gib if my integral seemed a little overwhelming, it's just one of the standard substitutions that I seem to remember. Perhaps if you told us your level we could set questions more appropriately.

    P.S. For my integral, a hyperbolic substitution would do the trick.
     
  14. Jan 2, 2007 #13

    Gib Z

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    Umm, Im not sure If i got it, But basically for ssd's one I let u= tan x, because his u^2=tan x didnt seem to work for me..Anyway I got it to [tex]\int sqrt{u} \cdot \frac {1}{1+u^2} du[/tex], which I applied integration by parts to. I didn't want to get confused between my u substitution and my u in by parts, so i changed the u to an x for my working in parts and I basically got [tex]\frac {2x^{3/2}}{3(x^2+1)} + \frac {8x^{7/2}}{21}[/tex] where x = tan x, I know i probably havent chosen x again...might confuse some people...but you get the point. And I've never done a hyperbolic substitution before, unless you mean [tex]\sin x = \frac {e^{ix} - e^{-ix}}{2}[/tex] which I got from Eulers Identity, looks like [tex]\sinh ix[/tex] to me. Please tell me im not right, because I'd still have trouble with that lol. I dont actually know my level, thats why im doing this thread. Thanks for all your help guys.

    EDIT: I just tried my result for ssd's one in mathematica, didnt turn out nicely..
     
    Last edited: Jan 2, 2007
  15. Jan 2, 2007 #14

    cristo

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    Like I said, try the substitution u2=tanx. Why didn't this work for you? Did you substitute for dx correctly? Note:

    [tex] 2u du=sec^2(x)dx \Rightarrow 2u du = (1+tan^2(x))dx \Rightarrow dx=\frac{2u}{1+u^4} du [/tex] Substitute this in and evaluate the integral.
     
  16. Jan 2, 2007 #15

    Gib Z

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    I dont seem to understand how you got the first part in your Note. [tex] u^2 = \tan x[/tex] Then [tex]\frac {d}{dx} u^2 = \sec^2 x[/tex] then [tex] d(u^2) = \sec^2 x dx[/tex]. I dont know how you got 2udu...

    Sorry for not being good at this guys, I can tell Cristo's a little pissed lol, sorry...
     
    Last edited: Jan 2, 2007
  17. Jan 2, 2007 #16

    Gib Z

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    Well from what you showed me, I would get [tex]\int u^{\frac{1}{2}} \frac {2u}{1+u^4} du[/tex]. Just abit of Intergration by parts and trig substitution would get me home free, yea?
     
  18. Jan 2, 2007 #17

    ssd

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    Hint:
    sqrt(tanx)= (1/2)[{sqrt(tanx) + sqrt(cotx)} + {sqrt(tanx) - sqrt(cotx)}]
    Consider the first part:
    sqrt(tanx) + sqrt(cotx)
    To integrate this put z= sinx-cosx , then use, 2sinx.cosx= 1-(sinx-cosx)^2.
    The second part follows in a similar manner.
    I have given the result of the first part in reply #26 of this thread.
     
    Last edited: Jan 2, 2007
  19. Jan 2, 2007 #18

    cristo

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    I'm not pissed; sorry if my post seemed a bit blunt! HAve you come across the chain rule for differentiation? It states that [tex] \frac{d}{dx}(f)=\frac{df}{du}\frac{du}{dx}.[/tex]

    So, [tex]\frac{d}{dx}u^2=\sec^2x \Rightarrow \frac{d(u^2)}{du}{\frac{du}{dx}=\sec^2x \Rightarrow 2u \frac{du}{dx}=\sec^2x \Rightarrow 2u du=\sec^2x dx [/tex]
     
  20. Jan 2, 2007 #19

    cristo

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    I think you have one mistake here, namely the [itex]\sqrt u[/itex]. Since [itex]u^2=tanx [/itex] this should simply be u. I don't think you need any more substitution or int. by parts- just use partial fractions.
     
  21. Jan 2, 2007 #20

    Gib Z

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    O CRAP! I am so so SO sorry! Mental Blank, yes im completely fine with differentiation. O thank you. Ok Well that ends up giving me [tex] 2\int \frac {u^2}{1+u^4} du[/tex] Umm...I don't get how you would do it with partial fractions, i was thinking of trig sub, i duno how though. What i did notice though, even though it probably doesnt help me, is letting u^2=tan(x/2) gives me [tex]\int \tan x du[/tex]..
     
    Last edited: Jan 2, 2007
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