# Homework Help: Integral sqrt((5-x)/x)

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1. Dec 31, 2017

### Vriska

1. The problem statement, all variables and given/known data
integration with respect to x

2. Relevant equations

integral 1/sqrt (a^2 - x^2) = arcsin(x/a)
3. The attempt at a solution
image attached, the arcsine term in 5/2 arcsin((2x-5)/5) it should be 5 arcsine(sqrt(x/5))

2. Dec 31, 2017

### Vriska

3. Dec 31, 2017

### Delta2

I think the arcsine formula for computing $\int \frac{1}{\sqrt{(\frac{5}{2})^2-(x-\frac{5}{2})^2}}dx$ cannot be applied because for x=0 the denominator goes to 0.

Last edited: Dec 31, 2017
4. Dec 31, 2017

### QuantumQuest

It would be far better to use Latex in order to be helped because checking through the attached image is somewhat hard.
Now, for the integral, I would hint in the way of substitutions with $u = \frac{5 - x}{x}$ being the most obvious. Have you tried this way?

5. Dec 31, 2017

### Ray Vickson

6. Dec 31, 2017

### vela

Staff Emeritus
When I plot those two functions in Mathematica, they appear to differ only by a constant, namely $5\pi/4$.

7. Dec 31, 2017

### vela

Staff Emeritus
You have the same issue for
$$\int \frac{dx}{\sqrt{1-x^2} } = \arcsin x + C$$ at $x=\pm 1$.

8. Dec 31, 2017

### Delta2

Well for some reason I thought the domain of the function should contain 0, but now I understand this is not necessary. Should just state that the result holds for $x\neq 0$

Last edited: Dec 31, 2017
9. Dec 31, 2017

### Vriska

Yeiks, they look so different though. looks like constants do indeed make a huuuge difference. Thank you though, this was giving me a headache.

10. Dec 31, 2017

### Vriska

That substitution isn't working for me :/, I'm getting $du =\frac{5}{x^2} dx$ which gives me $dx= \frac{1}{(u^2+1)^2}$ integral is this times a root u

11. Jan 1, 2018

### ehild

Try the substitution $u=\sqrt{\frac{5-x}{x}}$

12. Jan 1, 2018

### Vriska

This is u^2 =previous substitution . I'd get integral mentioned in previous reply but instead of multiplied by root u it'd be divided by 2 sqrt(u). any pointers on how i might use this more meaningfully?

13. Jan 1, 2018

### ehild

What is x if $u^2=\frac{5-x}{x}$ and what is dx? Is there any square root in the integrand?

14. Jan 1, 2018

### PeroK

If you have an answer to an integral that you are not sure about, you can always differentiate it to see whether you get back the integrand.

15. Jan 1, 2018

### epenguin

If you have any textbook with a chapter or two on integration, you will find a list related things, practically a family of them, there.
In this family it is quite common to have integrations that work only within certain range of x, a different formula outside it. This can correspond to something physical such as existence of escape velocity or runaway reaction outside the ‘tame’ region.

Not everybody does it this way but I recommend to change the variable so as not to have the constant $a$ within the formula to be integrated, e.g let $x = 5aX$, a new variable . That way you can get integrands that are more recognisable and reduced to a smaller standard set. (You cannot forget this original $a$, especially when you also have to change the $dx$ and also at the end of the calculation .)

There are then various ways, but I think what you were doing looks unnecessarily complicated than the simplest approach is just a further change of variable defining a $y = sin X$

As things like this do not usually come out of the blue, I wonder if this reminds you of anything in that lesson or book?

16. Jan 1, 2018

### stevendaryl

Staff Emeritus
A general trick is to try to get rid of the square root by making a variable substitution. In the case of $\sqrt{\frac{5-x}{x}}$, if you let $x = 5 cos^2(\theta)$, then this becomes: $\sqrt{\frac{1-cos^2(\theta)}{cos^2(\theta)}} = \frac{sin(\theta)}{\cos(\theta)}$

17. Jan 1, 2018

### QuantumQuest

The substitution works but if you find it difficult to follow then try what ehild suggests in post #11. If you find $dx$ and substitute in the original integral for the new variable you'll get a manageable integral.

18. Jan 2, 2018

### Vriska

integral is $- \frac{2u^2du}{(u^2+1)^2}$ where u is what you suggested. This is to be followed by a substitution of u = tan z?

Last edited: Jan 2, 2018
19. Jan 2, 2018

### ehild

I used integration by parts . $f '= \frac{2u}{(u^2+1)^2}$, g=u.

Last edited: Jan 2, 2018