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 May 16th, 2007, 04:43 AM #1 Newbie   Joined: May 2007 From: Belgium Posts: 4 Thanks: 0 Integration problems hi, I can't seem to find the solution to these exercises 1) INT (cos(x))^7 dx 2) INT (x/(1+x^6)) dx (I tried to do something with tan^-1(x) but no success) thanks
 May 16th, 2007, 09:33 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 Re: Integration problems 1) ?(cos x)^7 dx Hint: use integration by parts, then replace sin²x with 1 - cos²x. 2) ?(x/(1+x^6)) dx Hint: 1 + x^6 ? (1 + x²)(1 - x² + x^4), so use carefully chosen partial fractions.
 May 16th, 2007, 02:35 PM #3 Newbie   Joined: May 2007 From: Belgium Posts: 4 Thanks: 0 thx, I tried to use those hints, maybe i've come a litle further now. I still can't see it tough hope you can read my writings (im quite bad in having some order :P ) 1) now i only seem to have gotten the same int + more 2) I get left with the circled equation for which I don't find a method to solve it
 May 16th, 2007, 06:51 PM #4 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Try trigonometric substitution on the second problem. Set up your triangle with a 1 on the bottom, an x^3 on the upright side, and a ?(1 + x^6) on the hypotenuse. The angle is in the lower left corner. x^3 = tan ? x = (tan ?)^(1/3) ?(1 + x^6) = sec ? 1 + x^6 = (sec ?)² x^3 = tan ? 3x² dx = (sec ?)² d? dx = (sec ?)² / 3x² d? dx = (sec ?)² / (3(tan ?)^(2/3)) d? Thus, ?x/(1+x^6) dx Is equivalent to: ?(tan ?)^(1/3) * (sec ?)²/ ((sec ?)² * (3(tan ?)^(2/3))) d? Cancel terms and pull out the constants: (1/3) * ?1/(tan ?)^(1/3) d? (1/3) * ?(cot ?)^(1/3) d? Well, that doesn't seem like quite so bad of a problem to integrate. See if you can integrate that, and then back-substitute to transform it into an answer in terms of 'x'.
 May 17th, 2007, 01:54 AM #5 Newbie   Joined: May 2007 From: Belgium Posts: 4 Thanks: 0 hi, thanks for your help, unfortunately I already tried that road and I too get left with the equation you write there, it seems to me its not so easy to solve at all If i write the integral in an integrator ( http://integrals.wolfram.com/index.jsp ) I get the following solution which looks prety hard to find, maybe with integration by parts, I'll keep trying
 May 17th, 2007, 04:40 AM #6 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Hmm... Have you tried playing with hyperbolic trig functions yet? It almost seems to me that they're somehow splitting up the original problem into a trig and a hyperbolic trig integral.
May 17th, 2007, 11:21 AM   #7
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Quote:
 Originally Posted by Kemblin now i only seem to have gotten the same int + more
You mean you found something like I = F - 6I and didn't notice that that implies I = F/7.

May 17th, 2007, 07:38 PM   #8
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Re: Integration problems

Quote:
 Originally Posted by Kemblin hi, I can't seem to find the solution to these exercises 1) INT (cos(x))^7 dx 2) INT (x/(1+x^6)) dx (I tried to do something with tan^-1(x) but no success) thanks
The first integral becomes INT (1-u^2)^3 du after the substitution u = sin(x), and the second becomes (1/2) INT du/(1+u^3) after the substitution u = x^2.

 May 18th, 2007, 03:52 AM #9 Newbie   Joined: May 2007 From: Belgium Posts: 4 Thanks: 0 thanks for all the help, I think I found the first one now , I'll try the second one this eve hope it's correct...
 May 22nd, 2007, 10:02 PM #10 Newbie   Joined: May 2007 Posts: 7 Thanks: 0 There is another way to solve the problem There is another way to solve the problem http://img408.imageshack.us/my.php?imag ... 007wd4.jpg

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