integration problem!

[chimera]

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?

[The Bob]

\int (\frac{dx}{3cosx - 4sinx})

Does this help???:

\int tanxdx = \int \frac{sinx}{cosx}dx

It sounds like a partial fractions to me.

The Bob (2004 ©)

[PBRMEASAP]

Yes, you have to use the half angle formulas (or is it double angle formula?) for sin x and cos x.

\cos^2{x} = \frac {1 + \cos{2x}}{2} \ \ \mbox{and} \ \ \sin^2{x} = \frac {1-\cos{2x}}{2}

[dextercioby]

It looks really ugly.

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

Make the substitution:

x=2\arctan u (<=> u=\tan\frac{x}{2}) (2)

,under which simple trigonometry and differentiation will show that

dx=\frac{2 du}{1+u^{2}} (3)

\sin x= \frac{2u}{1+u^{2}} (4)

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

Can u continue from here...?

Daniel.

[chimera]

thanks, i got it =)

[One-D]

Dextercioby, I don't get what u wrote. in the 4th warning. the (2) would u tell me. thanx

[arildno]

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.

[dextercioby]

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

Daniel.

[arildno]

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

Daniel.
I already knew you were a good and inoffensive boy..:wink:

[dextercioby]

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.

[One-D]

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