# Integration problem!

1. Mar 7, 2005

### 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?

2. Mar 7, 2005

### 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.

3. Mar 7, 2005

### PBRMEASAP

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

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

4. Mar 7, 2005

### 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.

5. Mar 7, 2005

### chimera

thanks, i got it =)

6. Mar 10, 2005

### One-D

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

7. Mar 10, 2005

### 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.

8. Mar 10, 2005

### dextercioby

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

Daniel.

9. Mar 10, 2005

### arildno

I already knew you were a good and inoffensive boy..

10. Mar 10, 2005

### dextercioby

Thanks for the trust.Marlon feels the same way,though i don't remember any warning gotten from the clashes we've had...

Daniel.

11. Mar 12, 2005

### One-D

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