- #1

- 2

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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?

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- Thread starter chimera
- Start date

- #1

- 2

- 0

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?

- #2

- 1,100

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Does this help???:

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

It sounds like a partial fractions to me.

The Bob (2004 ©)

- #3

- 191

- 2

[tex]\cos^2{x} = \frac {1 + \cos{2x}}{2} \ \ \mbox{and} \ \ \sin^2{x} = \frac {1-\cos{2x}}{2}[/tex]

- #4

- 13,077

- 640

[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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Daniel made a very common and useful change of variables.

That's all there is to it.

- #8

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Incidentally i have 4 warnings...:rofl: :uhh:

Daniel.

Daniel.

- #9

arildno

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I already knew you were a good and inoffensive boy..dextercioby said:Incidentally i have 4 warnings...:rofl: :uhh:

Daniel.

- #10

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Daniel.

- #11

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thanx. know i understand. it's only a simple subs. thx anyway.

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