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Integral, hint

  1. Dec 28, 2007 #1
    Integral, hint plz

    [tex]\int\frac{\cos^{\frac{5}{2}}x}{\sqrt{\sin{x}}}dx[/tex]

    Tried: splitting it up, trig identities = no go!

    [tex]\int\frac{\cos{x}\cos^{\frac{3}{2}}x}{\sqrt{\sin{x}}}dx[/tex]

    Broke up cosine cubed then ended up realizing I would have a cosine left over so that wasn't a good idea.

    [tex]\int\cos^{2}x\sqrt{\cot{x}}dx[/tex]

    Then my world crumbled.
     
  2. jcsd
  3. Dec 28, 2007 #2
    Try the transformation

    [tex] x=\arctan(t^2), \quad d\,x=\frac{2\,t}{1+t^4}\,d\,t [/tex]
     
  4. Jan 11, 2008 #3
    @ rocophysics: Have you been able to solve this integral? Do you need any information, because it is quite involved and won't post the method if not necessary. It will take me about an hour or so to type everything :-)
     
  5. Jan 11, 2008 #4
    Lol, can I get a hint? I wasn't sure how to apply Rainbow Child's substitution so I decided to put it off for a while. :p
     
  6. Jan 11, 2008 #5
    Applying the transformation I wrote, you get

    [tex]I=\int\frac{2}{(1+t^4)^2}\,d\,t[/tex]

    since

    [tex] \cos x=\frac{1}{\sqrt{1+\tan^2 x}}\Rightarrow \cos x=\frac{1}{\sqrt{1+t^4}}, \quad \sin x=\sqrt{1-\cos^2x}\Rightarrow \sin x=\frac{t^2}{\sqrt{1+t^4}}[/tex]

    and I think from here is a easy task! :smile:

    P.S. What about the books, roco?
     
  7. Jan 11, 2008 #6
    OK, the method that rainbow child proposed was correct. You need to set
    [tex]t^2=tan(x)[/tex]
    giving
    [tex]dx=\frac{2t}{1+t^4}dt[/tex]
    Putting this into the integral gives then
    [tex]I=2 \cdot \int \frac{dt}{(1+t^4)^2}[/tex]
    This one now needs to be solved by expanding the thing into the following partial fraction equation:
    [tex]\frac{1}{(t^4+1)^2}=\frac{1}{(t^2-\sqrt{2}t+1)^2 \cdot (t^2+\sqrt{2}t+1)^2} = \frac{At+B}{t^2-\sqrt{2}t+1} + \frac{Ct+D}{(t^2-\sqrt{2}t+1)^2} + \frac{Et+F}{t^2+\sqrt{2}t+1} + \frac{Gt+H}{(t^2+\sqrt{2}t+1)^2}[/tex]
    Solving this is now the tedious step :-) You get finally:
    [tex]A=-\frac{3\sqrt{2}}{16}[/tex]
    [tex]B=\frac{3}{8}[/tex]
    [tex]C=-\frac{\sqrt{2}}{8}[/tex]
    [tex]D=\frac{1}{8}[/tex]
    [tex]E=\frac{3\sqrt{2}}{16}[/tex]
    [tex]F=\frac{3}{8}[/tex]
    [tex]G=\frac{\sqrt{2}}{8}[/tex]
    [tex]H=\frac{1}{8}[/tex]
    The remaining integrals are fairly standard but still require a bit of work. Hope this helps, if anything is unclear, let me know.
     
  8. Jan 12, 2008 #7
    Amazing, ok let me try that substitution and see how far I can get w/o looking at your soln.

    @Rainbow Child, I really appreciate your generosity :-]]] Thanks.
     
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