Efficient Trig Substitution for Simple Integration Problem

[dtl42]

[SOLVED] Simple Integration Problem

Homework Statement

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

Homework Equations

...none

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.

 

[rocomath]

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

Yes?

 

[dtl42]

Yea, sorry about that.

 

[dtl42]

OK, I updated it with the Latex.

 

[dtl42]

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

 

[rocomath]

Grr. I'm stumped. Must think harder.

 

[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...

 

[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.

 

[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.

 

[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? :)

 

[rocomath]

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

 

[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?

 

[Rainbow Child]

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

 

[Rainbow Child]

Oupps! VietDao29 was faster

 

[VietDao29]

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

Yah, thanks a lot..

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?

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..

 

[rocomath]

What situations does it work for?

When all hope is lost :-D

 

[dtl42]

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

 

[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.