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

Yea, sorry about that.

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)

So, your integral will become:

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

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.