# Simple Integration Problem

1. Feb 22, 2008

### dtl42

[SOLVED] Simple Integration Problem

1. The problem statement, all variables and given/known data
Integrate: $$\int\frac{dx}{3\sin x+2}$$

2. Relevant equations
....none

3. The attempt at a solution

I tried to do this by u-sub, and that was unsuccessful, then after that I tried a trig. sub with 3sinx=2tan(theta)^2, and that got very messy, really fast. So, I'm kinda stuck, any help would be appreciated.

Last edited: Feb 22, 2008
2. Feb 22, 2008

### rocomath

$$\int\frac{dx}{3\sin x+2}$$

Yes?

3. Feb 22, 2008

### dtl42

4. Feb 22, 2008

### dtl42

OK, I updated it with the Latex.

5. Feb 22, 2008

### dtl42

do you mean : $$dt=\frac{dx}{\sqrt{x^2-1}}$$ ?

6. Feb 22, 2008

### rocomath

Grr. I'm stumped. Must think harder.

7. Feb 22, 2008

### dtl42

Yea, I've been working on and off with this one for a few hours, and I even plugged it into Mathematica's Integrator, but I get some crazily complicated stuff. I'm really in need of help with it...

8. Feb 22, 2008

### rocomath

Well, I have this so far ...

$$t=\sin x$$
$$x=\sin^{-1}t$$
$$dx=\frac{dt}{\sqrt{1-t^2}}$$

$$\int\frac{dt}{(3t+2)\sqrt{1-t^2}}$$

Looking at Mathematica, I may be on the right track. I'm going to try a hyperbolic substitution now.

9. Feb 22, 2008

### rocomath

I think I'm going to start over, I need to use the so called world's sneakiest substitution, lol. If I can remember how to use it :p Gotta look through my book.

10. Feb 22, 2008

### VietDao29

When seeing some integrals like this, i.e, those trigonometric functions, that clearly cannot use other normal u-substitution, then, we must think about substituting tan(x/2).

So, let:

• u = tan(x/2).

~~~> du = (1/2) (1 + tan2(x/2)) dx = [1/2 (1 + u2)] dx

~~~> dx = (2 du) / (1 + u2)
• sin(x) = (2u) / (1 + u2)

You know this formula, right? If not, here's a simple proof:

$$\sin (x) = \sin (2 \times \frac{x}{2}) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)}$$

$$= 2 \frac{1}{\frac{1}{\cos ^ 2 \left( \frac{x}{2} \right) }} \tan \left( \frac{x}{2} \right) = 2 \frac{\tan \left( \frac{x}{2} \right)}{1 + \tan ^ 2 \left( \frac{x}{2} \right) } = \frac{2u}{1 + u ^ 2}$$ (Q.E.D)

$$\int \frac{\frac{2}{1 + u ^ 2}}{3 \times \frac{2u}{1 + u ^ 2} + 2} du$$

Can you go from here? :)

Last edited: Feb 22, 2008
11. Feb 22, 2008

### rocomath

Very nice VietDao :-] I was lookin through my book and next refresh I see your work. Very clean!

12. Feb 22, 2008

### dtl42

Thanks a lot, that is the first time that I've seen that substitution used. How often is it usually employed? What situations does it work for?

13. Feb 22, 2008

### Rainbow Child

Try this
$$x=2\,\arctan u\Rightarrow u=\tan \frac{x}{2},\,d\,x=\frac{2\,du}{1+u^2}$$
with
$$\sin x=\frac{2\,\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\Rightarrow \sin x=\frac{2\,u}{1+u^2}$$

14. Feb 22, 2008

### Rainbow Child

Oupps! VietDao29 was faster

15. Feb 22, 2008

### VietDao29

Yah, thanx a lot..

Well, it works for some types like:

$$\int \frac{\alpha}{\beta \sin (x) + \gamma } dx$$, $$\int \frac{\alpha}{\beta \cos (x) + \gamma } dx$$, $$\int \frac{\alpha}{\beta \sin (x) + \gamma \cos (x) + \delta } dx$$, ...

where alpha, beta, gamma, and delta are all constants. :)

There are some more, but the 3 listed above are the most common ones..

Last edited: Feb 22, 2008
16. Feb 22, 2008

### rocomath

When all hope is lost :-D

17. Feb 22, 2008

### dtl42

Oh, that's interesting, it seems I've been deprived of any exposure to those types. Thanks to everyone for answering so quickly.

18. Feb 22, 2008

### Gib Z

It reduces any rational function of sin x and cos x into a rational function of t. In fact its not when all hope is lost, its the first thing you do for many trig integrals.