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Goniometric Integral Problem

  1. Jul 3, 2005 #1

    TD

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    Hi,

    I'm having some trouble with solving this indefinite integral.

    [tex]
    \int {\sqrt {\frac{{6\cos ^2 x + \sin x\cos (2x) + \sin x}}{{2 - \sin x}}} } dx
    [/tex]

    I was able to lose the sin(x) and get a cos(x) out of the square root by doing this:

    [tex]
    \int {\sqrt {\frac{{6\cos ^2 x + \sin x(2\cos^2 x -1) + \sin x}}{{2 - \sin x}}} } dx
    [/tex]

    [tex]
    \int {\sqrt {\frac{{6\cos ^2 x + 2\sin x\cos^2 x -\sin x + \sin x}}{{2 - \sin x}}} } dx
    [/tex]

    [tex]
    \int {\sqrt {\frac{{\cos ^2 x(6 + 2\sin x)}}{{2 - \sin x}}} } dx
    [/tex]

    [tex]
    \int \cos x{\sqrt {\frac{{6 + 2\sin x}}{{2 - \sin x}}} } dx
    [/tex]

    Then I did a substitution, [tex]y = \sin x \Leftrightarrow dy = \cos xdx[/tex] to get:

    [tex]
    \int {\sqrt {\frac{{6 + 2y}}{{2 - y}}} } dy
    [/tex]

    I think I can say this looks a lot better than the initial integral, but after that I used about 3 more substitutions and 2 sides of paper. I finally got something but it wasn't all correct I'm affraid.
    I was wondering if I started out wrong or if someone sees an easy way to continue (or an easier to start).

    Naughty integral if you ask me :bugeye:
     
    Last edited: Jul 3, 2005
  2. jcsd
  3. Jul 3, 2005 #2

    dextercioby

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    It looks okay so far.Now try the sub

    [tex]\frac{6+2y}{2-y}=t^{2} [/tex]

    Daniel.
     
  4. Jul 3, 2005 #3

    TD

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    Hi, perhaps I should've went on but that's indeed what I did first.

    I then got:

    [tex]
    \int {\frac{{20t^2}}{{(t^2+2)^2}} } dt
    [/tex]

    I was able to expand that to:

    [tex]
    \int {\frac{{20}}{{t^2+2}} - \frac{{40}}{{(t^2+2)^2}}} dt
    [/tex]

    The first one is no problem, I can easily get there with a ArcTan like this:

    [tex]
    \int {\frac{{20}}{{t^2+2}}} dt = 10\sqrt 2 \arctan \left( {\frac{{\sqrt 2 t}}{2}} \right) (+ C)
    [/tex]

    But what about the second one? I know it gived an ArcTan as well, but not only that. I've seen recursion formula's for it but is there a way to do it 'by hand'?

    PS: I read you're temporarily in Leuven, I study in Brussels - that's very nearby :biggrin:
     
  5. Jul 3, 2005 #4

    dextercioby

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    U could have easily done it using part integration

    [tex] 10\int t \ \frac{d\left(t^{2}+2\right)}{\left(t^{2}+2\right)^{2}} =-10\frac{t}{t^{2}+2} +10\int \frac{dt}{t^{2}+2} [/tex]

    Can you take it from here...?

    Daniel.
     
    Last edited: Jul 3, 2005
  6. Jul 3, 2005 #5

    TD

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    Of course! Then I get:

    [tex]
    \int {\frac{{20t^2}}{{(t^2+2)^2}} } dt = 5\sqrt 2 \arctan \left( {\frac{{\sqrt 2 t}}{2}} \right) - \frac{{10t}}{{t^2 + 2}} + C
    [/tex]

    Susbstituting back:

    [tex]\frac{6+2y}{2-y}=t^{2} \Leftrightarrow t=\sqrt{\frac{6+2y}{2-y}} \,\,\, gives:[/tex]

    [tex]
    5\sqrt 2 \arctan \left( {\sqrt {\frac{{y + 3}}{{2 - y}}} } \right) + \sqrt 2 \left( {y - 2} \right)\sqrt {\frac{{y + 3}}{{2 - y}}} + C
    [/tex]

    Susbstituting back:

    [tex]y=\sin x \,\,\, gives:[/tex]

    [tex]
    5\sqrt 2 \arctan \left( {\sqrt {\frac{{\sin x + 3}}{{2 - \sin x}}} } \right) + \sqrt 2 \left( {\sin x - 2} \right)\sqrt {\frac{{\sin x + 3}}{{2 - \sin x}}} + C
    [/tex]

    I hope that's it. Finally, thanks :smile:
     
  7. Jul 3, 2005 #6

    Zurtex

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    Yep, took a bit of messing to check it was the right, but it is.
     
  8. Jul 3, 2005 #7

    TD

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    Yes, I had Mathematica take the derivative and after FullSimplify, the initial expression almost identically rolled out. Just had to put the cos (and a factor) back in the square root :smile:
     
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