# Trigonometric substitutions

1. Nov 5, 2012

### MarcZZ

1. The problem statement, all variables and given/known data

Hi I need help with the following integral.

$\int_0^2 \! \frac{1}{(x^2+4)} \, dx$

2. Relevant equations

I believe that these are both trigonometric substitutions. However, these are the simplest in my textbook and I can't even understand them. :-(

3. The attempt at a solution

a) I said t = 4x

Thus $\frac{1}{4} \int_0^2 \! \frac{1}{(x^2+1)} \, dt$

So dt = 4dx

$\frac{1}{4} \int_0^2 \! \frac{4}{(x^2+1)} \, dx$

Therefore...

1/4 (tan^-1(x))|2 = b and 0 = a
1/4 (((tan^-1(1(2)/4)) - (tan^-1(1(0)/4)))

Somehow I am supposed to get pi/8 but I don't understand the math once I get down to this point...

Am I doing this all wrong? Thanks ahead for any help... :)

2. Nov 5, 2012

### Mukuro

Re: Integral...

Just using u-substitution (in this case t), won't accomplish much, I believe the problem calls for strict memorization of how the integrals of these kind of functions works.

We know that, the integral of 1/(x^2+1) is tan inverse of x, however a general way to look at it is [ (sqrt S) * tan inverse of (sqrt S * x) / S ) ] / S

Where S = the constant added int he denominator of the derivative of tan inverse.

Thus since you have 4, the correct integral is [2 * tan inverse (2*x/4) ] / 4

which turns out to be [ tan inverse (x/2) ] / 2, this is the correct integral and should get you yur answer

3. Nov 5, 2012

### MarcZZ

Re: Integral...

And where could I learn this, as I sure as heck don't have anything on this in my notes or textbook...

4. Nov 5, 2012

### Mukuro

Re: Integral...

Well I happen to be a student still, but when I was taking calculus, tbh my teacher just taught me how it worked. She didn't even bother to explain it, she just gave us the fact that the integral of that type of function was tan inverse of x and showed us how it worked.
I do not find that books cover every single little things, and if they happen to do they probably don't make it easy.

5. Nov 5, 2012

### MarcZZ

Re: Integral...

I understand that 1 / 1 + x^2 is tan^-1(x), however I don't see where you got all these other (sqrt S) tan inverse (sqrt S * x) and so forth.

6. Nov 5, 2012

### Mukuro

Re: Integral...

It's just how the integration of the functions works. The constant added in the denominator will ALWAYS follow the pattern I typed out. For instance, the integral of 1/ x^2 + 3

will be,

[sqrt 3 * tan inverse (sqrt 3 * x/3)]/3 this can be shown using something on the internet or a graphing calculator.

7. Nov 5, 2012

### Zondrina

Re: Integral...

Yes the function you have in question has an elementary anti derivative. Namely arctan(x).

What is the derivative of arctan(x)?

8. Nov 5, 2012

### SammyS

Staff Emeritus
Re: Integral...

Furthermore, you should use the substitution, x = 2t. That way, x2 = 4t2 .

9. Nov 6, 2012

### HallsofIvy

Staff Emeritus
Re: Integral...

Well, this is incorrect because you have "dt" but still have "x" in the integrand.

And now all you have done is write almost the original integral except with a "1/4" in front of it! It cannot possibly be the same thing.

If you want to make the substitution t= 4x, then do the substitution! x= t/4 so, yes, dx= dt/4. And x^2+ 4= t^2/16+ 4. Surely, that's not what you wanted? That's much more complicated than what you have originally!

I think you were trying to get rid of the "4" in the denominator: you want $x^2+ 4= 4t^2+ 4= 4(t^2+ 1)$ so the substitution you want is x= 2t, NOT "t= 4x". With x= 2t, dx= 2dt and $x^2+ 4= 4t^2+ 4= 4(t^2+ 1)$ so the integral becomes
$$\int \frac{2dt}{4(t^2+ 1)}= \frac{2}{4}\int\frac{dt}{t^2+ 1}= \frac{1}{2}\int\frac{dt}{t^2+ 1}$$

But don't forget to change the limits of integration. Originally they are x= 0 and x= 2. With x= 2t= 0, t= 0 and with x= 2t= 2, t= 1 so your integral should be
$$\frac{1}{2}\int_0^1 \frac{dt}{t^2+ 1}$$

Yes, you are doing that all wrong. You appear to be trying to copy half remembered examples blindly without thinking about what you are doing.