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Proof of integrals

  1. Nov 30, 2006 #1
    I have just been shown integrals without proof and would rather understand where they came from. I'm sure these have been posted somewhere on here but dont know what to search for. If a website have all these proofs Ill gladly look at them.

    \int \frac {1} {(x^2 - a^2)^{0.5}} = ln(x + (x^2 - a^2)^{0.5})

    \int \frac {1} {x^2 + a^2} = \frac {1} {a} tan^{-1}( \frac {x} {a} )

    Last edited: Nov 30, 2006
  2. jcsd
  3. Nov 30, 2006 #2
    1. Let [tex] x = a\sec \theta [/tex]

    2. Let [tex] x = a\tan \theta [/tex]
  4. Nov 30, 2006 #3


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    Well, try some trigonometric substitution and see where it gets you. I think you can use [tex]x = a \sec \theta[/tex] for the first integral.

    Edit: oops, too late.
  5. Nov 30, 2006 #4


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    For the first instead of sec, you might try [itex] x=a\sinh t [/itex].

  6. Nov 30, 2006 #5
    Why are trigonometry substitutions the key, do I just need to accept that?
  7. Nov 30, 2006 #6


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    Trigonometric substitutions are just special cases of the technique of integration by substitution. The idea behind integration is that you chose a substitution that would simplify your integral to such an extent that you can perform the integral; it is equivalent to the chain rule in differentiation. It simply turns out that trigonometric substitutions are particularly good at simplifying integrals. For example if you had something of the form;

    [tex]I = \int \frac{dx}{\sqrt{a-ax^2}}[/tex]

    You would use the substitution [itex]x = \sin\theta[/itex] since when you substitute you would obtain something significantly simpler, can you see why?. Once you've done a few integrals, you get a feel for which substitutions to use.
    Last edited: Nov 30, 2006
  8. Nov 30, 2006 #7
    number 2

    x = atan \theta , a^2 + x^2 = a^2 + a^2 tan^2 \theta

    common factor of a^2, a^2(1 + tan^2 \theta)

    is there an identity for this?
    Last edited: Nov 30, 2006
  9. Nov 30, 2006 #8


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    Yes, there is as a matter of fact;

    [tex]1+\tan^2\theta = \sec^2\theta[/tex]
  10. Nov 30, 2006 #9


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    From [itex]sin^2 \theta + cos^2 \theta= 1[/itex] we can derive:

    [tex]cos^2 \theta= 1- sin^2 \theta[/tex]
    [tex]sec^2 \theta= 1+ tan^2 \theta[/tex]
    [tex]tan^2 \theta= sec^2 \theta - 1[/tex]
    So if you have something involving
    [tex]\sqrt{1- x^2}[/tex]
    [tex]\sqrt{1+ x^2}[/tex]
    [tex]\sqrt{x^2- 1}[/tex]
    The substitution [itex]sin \theta= x[/itex], [itex]tan \theta= x[/itex], or [itex]sec \theta = x[/itex] will put a single square inside the square root, getting rid of the square root.

    As dextercioby pointed out, for hyperbolic functions, we have the identity [itex]cosh^2 x- sinh^2 x= 1[/itex] so
    [tex]cosh^2 x= 1+ sinh^2 x[/tex]
    [tex]sech^2 x= tanh^2- 1 x[/tex]
    [tex]tanh^2 x= 1- sech^2 x[/tex]
    so the hyperbolic functions can be used similarly.
  11. Nov 30, 2006 #10


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    Always write 'dx' in your integral. It made you forget to differentiate the substitited x = ... and plug it into the integral.
  12. Nov 30, 2006 #11
    The previewing drives me up the wall :[

    [tex]\int \frac {1} {{(x^2 - a^2)}^{0.5}} sub-x^2=a^2sec^2 \theta [/tex]

    [tex] \int \frac {1} {((a^2sec^2 \theta) - a^2)^{0.5}} [/tex]

    [tex] \int \frac {1} {(a^2(sec^2 \theta - 1))^{0.5}} [/tex]

    [tex] \int \frac {1} {(a^2 tan^2 \theta)^{0.5}}[/tex]

    I assume the square and square root cancel out, but dont know where to go after this, I am guessin integrate?
  13. Nov 30, 2006 #12


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    Once again, your integrals without the differential form 'dx' represent nothing.
  14. Nov 30, 2006 #13
    You end up with [tex] \frac{1}{a}\int \frac{1}{\tan\theta} a\sec \theta \tan \theta [/tex]
  15. Nov 30, 2006 #14
    I cannot see how to go from mine to yours courtrigrad, I understand the 1/a but not the rest
  16. Nov 30, 2006 #15
    [tex] \int \frac{1}{\sqrt{a^{2}\tan^{2}\theta}}\; dx = \int \frac{1}{a\tan \theta}\; dx = \frac{1}{a} \int \frac{1}{\tan\theta} \; dx [/tex]

    where [tex] dx = a\sec \theta \tan \theta [/tex]
  17. Dec 4, 2006 #16

    Gib Z

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    You don't need trig sub for these, there's a much easier way for say, number 2. Remember he wants a proof, not a derivation. Just differentiate the right side, and if you get the same integral it's proven.

    [tex]y= arc tan (x)[/tex]
    then [tex]x=tan (y)[/tex]
    [tex] \frac {dx}{dy} = sec^2 y[/tex] (proven by using quotient rule, letting u= sin y and v= cos y)
    [tex](sec^2 y) = 1 + tan^2 y[/tex]
    [tex] \frac {dx}{dy} = 1 + tan^2 y[/tex]

    Remember [tex]tan (y) = x[/tex]

    [tex] \frac {dx}{dy} = 1 + x^2[/tex] then reciprocal both sides

    Therefore [tex] \frac {d}{dx} arc tan (x) = \frac {1}{1+x^2}[/tex]

    Now that we know that, we shall prove your integral with The Fundamental theorem of Calculus, damn i love it.

    \int \frac {1} {x^2 + a^2} dx

    From the denominator, take out the factor of [tex]a^2[/tex]. Now it's:

    \int \frac {1} {a^2 (1+\frac {x} {a})^2} dx

    We can take out the [tex] \frac {1}{a^2}[/tex] but just take out [tex]\frac {1}{a}[/tex] and leave the other [tex]\frac {1}{a}[/tex] inside, trust me.

    Now, inside the integral we have [tex]\frac {x}{a}[/tex]. Let it equal [tex]u[/tex]

    [tex]u= \frac {x}{a}[/tex]
    [tex]\frac {du}{dx} = \frac {1}{a}[/tex]

    Sub the [tex]\frac {du}{dx}[/tex] for [tex]\frac {1}{a}[/tex] and the [tex]u[/tex] for [tex]\frac {x}{a}[/tex] in our previous integral, we get:

    [tex]\frac {1}{a} \int \frac {1}{1+u^2} \frac {du}{dx} dx[/tex]

    The [tex]dx[/tex] 's cancel out, so we are left with:
    [tex]\frac {1}{a} \int \frac {1}{1+u^2} du[/tex]

    We know already that [tex]\int \frac {1}{1+u^2} du= arc tan (u) + C[/tex], so we solve and get:

    [tex]\frac {1}{a} arc tan (u) + C[/tex]

    Sub back in [tex]\frac {x}{a}[/tex] for [tex]u[/tex], we finally prove that your second integral equals :

    [tex]\frac {1}{a} arc tan (\frac {x}{a}) + C[/tex] as required. Hope i helped. Btw I finally got the hang of laTex :D
    Last edited: Dec 4, 2006
  18. Dec 4, 2006 #17
    That couldn't have answer my question better thanks Gib, now it just need to sink in for me.
  19. Dec 4, 2006 #18

    Gib Z

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    No Problem mate, any more help i gtg now but ill be back tomoro.


    well actually I would just take out the [itex]a^\frac {-1}{2}[/itex] to leave me with an integral I know is [itex]arcsin (x) + C[/itex]. Took what, 5 seconds? Much simpler if you remember a few integrals. O btw if anyone is wondering, what I just said can be proven in a similar fashion to what i just did a second ago, by changing [itex]arcsin (x) = y[/itex] to [itex]\sin (x) = y[/itex], differentiating, finding the recipricol. aww heck ill just do it here, but wont both with tex...

    y= arc sin x
    x= sin y

    dx/dy= cos y (hopefully you know this ones proof at least..)

    Now since we only define arc sin in the domain -pi/2 to pi/2 (otherwise with no domain there are an infinite number of solutions for each valid x).

    Anyway you may recognize to domain i set to be the 1st and 4th quadrants, where cos is positive. In these quadrants, cos y = +(1-sin^2 y)^0.5.

    remember sin y = x? cos y= (1-x^2)^0.5

    recipricol to be dy/dx = 1/ (1-x^2)^0.5 , as required.
    Last edited: Dec 4, 2006
  20. Dec 4, 2006 #19
    nice job Gib Z.
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