Integration problem!

  • Thread starter chimera
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  • #1
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could you please give me any idea to solve the problem below;

integral( dx / (3cosx-4sinx) )


and given a hint to make a subtitution u=tan(x/2), i've tried to write cosx and sinx in the form of cos (x/2) and sin(x/2), but it's seems like i'm not going anywhere, any suggestions?
 

Answers and Replies

  • #2
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[tex]\int (\frac{dx}{3cosx - 4sinx})[/tex]

Does this help???:

[tex]\int tanxdx = \int \frac{sinx}{cosx}dx[/tex]

It sounds like a partial fractions to me.

The Bob (2004 ©)
 
  • #3
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Yes, you have to use the half angle formulas (or is it double angle formula?) for sin x and cos x.

[tex]\cos^2{x} = \frac {1 + \cos{2x}}{2} \ \ \mbox{and} \ \ \sin^2{x} = \frac {1-\cos{2x}}{2}[/tex]
 
  • #4
dextercioby
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It looks really ugly.

[tex] I=:\int \frac{dx}{3\cos x-4\sin x} [/tex](1)

Make the substitution:

[tex] x=2\arctan u (<=> u=\tan\frac{x}{2}) [/tex] (2)

,under which simple trigonometry and differentiation will show that

[tex] dx=\frac{2 du}{1+u^{2}} [/tex] (3)

[tex] \sin x= \frac{2u}{1+u^{2}} [/tex] (4)

[tex] \cos x=\frac{1-u^{2}}{1+u^{2}} [/tex] (5)

Can u continue from here...?

Daniel.
 
  • #5
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thanks, i got it =)
 
  • #6
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Dextercioby, I don't get what u wrote. in the 4th warning. the (2) would u tell me. thanx
 
  • #7
arildno
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One-D: It's 4'th POST, not WARNING!
Daniel made a very common and useful change of variables.
That's all there is to it.
 
  • #8
dextercioby
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Incidentally i have 4 warnings...:rofl: :uhh:

Daniel.
 
  • #9
arildno
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dextercioby said:
Incidentally i have 4 warnings...:rofl: :uhh:

Daniel.
I already knew you were a good and inoffensive boy..:wink:
 
  • #10
dextercioby
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Thanks for the trust.Marlon feels the same way,though i don't remember any warning gotten from the clashes we've had...:wink:

Daniel.
 
  • #11
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thanx. know i understand. it's only a simple subs. thx anyway.
 

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