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May 16th, 2007, 04:43 AM   #1
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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
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May 16th, 2007, 09:33 AM   #2
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Re: Integration problems

1) ?(cos x)^7 dx
Hint: use integration by parts, then replace sinx with 1 - cosx.

2) ?(x/(1+x^6)) dx
Hint: 1 + x^6 ? (1 + x)(1 - x + x^4), so use carefully chosen partial fractions.
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May 16th, 2007, 02:35 PM   #3
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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
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May 16th, 2007, 06:51 PM   #4
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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 ?)

x^3 = tan ?
3x dx = (sec ?) d?
dx = (sec ?) / 3x d?
dx = (sec ?) / (3(tan ?)^(2/3)) d?


Thus, ?x/(1+x^6) dx

Is equivalent to:

?(tan ?)^(1/3) * (sec ?)/ ((sec ?) * (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'.
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May 17th, 2007, 01:54 AM   #5
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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
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May 17th, 2007, 04:40 AM   #6
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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.
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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.
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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.
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May 18th, 2007, 03:52 AM   #9
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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...
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May 22nd, 2007, 10:02 PM   #10
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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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