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Stuck on this integral

  1. Jan 17, 2005 #1
    integral of 1/(cscx-1)


    i can't think of any trig subs or anything......natural log prob? ;)....i udnno i'm truely clueless. all i need is a hint or more if u like ;) :wink:








    also, am i using integrals for this prob:

    Compute the average value of f(x)=x/(x+3) over the interval [-a,a], 0<a<3

    I think i will just get an expression for the answer
     
    Last edited: Jan 17, 2005
  2. jcsd
  3. Jan 17, 2005 #2
    Here is what I have done. So you should go through my working to check for errors.

    [tex]
    \int {\frac{1}{{\cos ecx - 1}}} dx
    [/tex]

    [tex]
    = \int {\frac{1}{{\left( {\frac{1}{{\sin x}} - 1} \right)}}} dx
    [/tex]

    [tex]
    = \int {\frac{1}{{\left( {\frac{{1 - \sin x}}{{\sin x}}} \right)}}} dx
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{1 - \sin x}}dx}
    [/tex]

    When I see something like the above I just think, something needs to be done to the denominator. I would use a technique similar to rationalising a surd expression as follows.

    [tex]
    = \int {\frac{{\sin x}}{{1 - \sin x}} \times \frac{{1 + \sin x}}{{1 + \sin x}}dx}
    [/tex]

    [tex]
    = \int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}dx}
    [/tex]

    Now use some trig identities as follows.

    [tex]
    = \int {\frac{{\sin x + \frac{1}{2} - \frac{1}{2}\cos 2x}}{{\cos ^2 x}}} dx
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{\cos 2x}}{{\cos ^2 x}}dx}
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{2\cos ^2 x - 1}}{{\cos ^2 x}}} dx
    [/tex]

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

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \int {dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} }
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\frac{1}{{\cos ^2 x}}} - \int {dx} }
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\sec ^2 x} dx - \int {dx} }
    [/tex]

    The first integral in the above line can be done by substitution or simply by inspection which can save time. :biggrin:

    [tex]
    = \frac{1}{{\cos x}} + \tan x - x + c
    [/tex]

    [tex]
    = \sec x + \tan x - x + c
    [/tex]
     
    Last edited: Jan 18, 2005
  4. Jan 18, 2005 #3

    dextercioby

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    Yes,Benny,good work... :smile: Though you could have shortcutted the part with the double angle on the cosine... :wink:

    Daniel.
     
  5. Jan 18, 2005 #4
    Thanks dextercioby. :smile:

    I see what you mean, I went around in circles. The question could have been finished as follows.

    [tex]
    \int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}} dx
    [/tex]

    [tex]
    = \int {\frac{{\sin x + \sin ^2 x}}{{\cos ^2 x}}} dx
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\tan ^2 x} dx}
    [/tex]

    [tex]
    = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\left( {\sec ^2 x - 1} \right)} dx}
    [/tex] The integral on the left can be integrated by substitution or using the fact that [tex]\frac{d}{{dx}}\left( {\sec x} \right) = \tan x\sec x[/tex]

    So [tex]\int {\frac{1}{{\cos ecx - 1}}dx} = \sec x + \tan x - x + c[/tex].
     
  6. Jan 18, 2005 #5

    dextercioby

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    I was thinking about something else...Right from the beginning...It doesn't really matter.It would have been another (more simple) trick besides the one you pulled...

    Daniel.

    P.S.These 2 tricks shorten the calculus,but the disadvantage is that u may not see either of them... :tongue2:
     
  7. Jan 18, 2005 #6
    In the numerator you could have used Sinx=-(-Sinx)=-(1-Sinx-1)
    which would have simplified the calc instead of *ing and /ing by 1+Sinx
     
    Last edited: Jan 18, 2005
  8. Jan 18, 2005 #7

    dextercioby

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    OMG,how can u say that?? :surprised Would u agree that what u said is basically an alambicated way of:
    [tex] \sin x=(\sin x-1)+1 [/tex]

    and the minus would appear naturally (when symplifying the denominator and the numerator),not forced...??

    Daniel;

    PS.Benny,that's the trick i mentioned...
     
  9. Jan 20, 2005 #8
    What do you mean by natural and forced??
    Yes I was talking about this only.[itex] \sin x=(\sin x-1)+1 [/itex]
    I added the -ve sign to prevent confusion,but it seems that it didn't work.
     
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