# Proof of an integral

1. Oct 31, 2007

### lazypast

hi, ive a function here and im finding it hard to integrate

$$\int \frac {1}{\sqrt{a^2 - x^2}}$$

let x = asin(theta) so $$\frac{dx}{d \theta}=acos \theta$$

$$dx=acos \theta d \theta$$

$$\int \frac {acos \theta d \theta}{\sqrt{a^2 - a^2 sin^2 \theta}}$$

$$\int \frac {acos \theta d \theta}{\sqrt{a^{2} (1 - sin^2 \theta)}}$$

$$\int \frac {acos \theta d \theta}{acos \theta} = \int d \theta$$

i cant continue this as i cant get the right equation which is able to be integrated. any ideas?

2. Oct 31, 2007

### d_leet

What is the problem you're having, you reduced a diffcult integral into a practically trivial one?

3. Oct 31, 2007

### lazypast

i reduced it to dtheta, but im unable to sub anything back into the integral which i can solve.

$$d \theta = \frac {dx}{acos \theta}$$

$$\int \frac {dx}{acos \theta}$$

but this isnt any more help to me, i know what the answr is ( sin^-1 (x/a)) but arent close to it

4. Oct 31, 2007

### d_leet

Don't substitute back yet, that defeats the whole point of teh substitution. Just integrate d(theta).

5. Oct 31, 2007

### TMM

And THEN substitute back in with a reference triangle or just theta in terms of x.

That's how trig substitution works.

6. Nov 1, 2007

### Gib Z

Would it help if you rewrote it as $$\int 1 d\theta$$? What (family of) function(s) whose derivative with respect to theta gives 1? As d_leet pointed out, you really already did the hard part, Made some nice soup but don't know how to use a spoon :(

7. Nov 1, 2007

### HallsofIvy

Staff Emeritus
This also has nothing to do with Differential Equations so I'm moving it to "Calculus and Analysis".

8. Nov 1, 2007

### lazypast

$$\int d \theta = \theta$$

$$\frac {dx}{d \theta} = acos \theta$$

$$\theta = cos^{-1}(\frac {1}{a} \frac {dx}{d \theta})$$

and this still has a dx/dtheta in it. i really am stumpted here

9. Nov 1, 2007

### lazypast

aha ive been abit blind.

$$x = asin \theta$$

$$\theta = sin^{-1} \frac {x}{a}$$

i dont understand the way you show it.
when f(x) = theta, f'(x) = 1 when you differentiate with respect to theta

thanks for the help anyway

10. Nov 2, 2007

### Gib Z

Are you familiar with the fundamental theorem on calculus?

11. Nov 2, 2007

### lazypast

i havent been tought the fundamental theory of calculus,but i can do integration and differentiation which is what i thought is basically was

12. Nov 2, 2007

### Gib Z

:( There are a few versions of the theorem, and they are all equivalent, you really need to know the theorem, and I'm sure you already do but just don't realize it.

It basically says that $$\int^b_a f(x) dx = F(b) - F(a) , \mbox{where} F'(x)=f(x)$$, or equivalently, $$\frac{d}{dx}\int^x_a f(t) dt = f(x)$$. Try and see why they are the same.

13. Nov 2, 2007

### ZioX

What's the derivative of theta with respect to theta?

14. Nov 2, 2007

### lazypast

derivative of theta with respect to theta id say is one.

after looking at some concepts of the fundamental theorm of calculus i cant seem to understand them. you say

$$\int^b_a f(x) dx = F(b) - F(a) , \mbox{where} Fsingle-quote(x)=f(x)$$

i would have thought F(x) = f ' (x)

i dont know why you use F and f

if you replace f(x) with y, then i suppose F(x) could be z?

15. Nov 2, 2007

### Gib Z

It doesn't matter what letters you use, the theorem stays the same though :( It is F'(x) = f(x), you really need to learn this, check your textbook.

16. Nov 3, 2007

### HallsofIvy

Staff Emeritus
These were posts 8 and 9 but we went on for 6 more posts anyway!

(Oops, 7 now.)

17. Nov 3, 2007

### Gib Z

Well its because we found a serious problem in his understanding :( He says he knows differential and integral calculus, yet has never heard of the Fundamental Theorem before!!!

18. Nov 5, 2007

### lazypast

i have heard of it just never been tought it. ive just found out that $$\int f(x) = F(x)$$ and that was the notation i didnt understand. like the capital F had been plucked from no where