Integration with Respect to x: Integral of sqrt((5-x)/x)

[Vriska]

Homework Statement

integration with respect to x

Homework Equations

integral 1/sqrt (a^2 - x^2) = arcsin(x/a)

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))

 

[Vriska]

 

[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.

 

[QuantumQuest]

Homework Statement

integration with respect to x

Homework Equations

integral 1/sqrt (a^2 - x^2) = arcsin(x/a)

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))

View attachment 217650

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?

 

[Ray Vickson]

Homework Statement

integration with respect to x

Homework Equations

integral 1/sqrt (a^2 - x^2) = arcsin(x/a)

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))

Your image is an unreadable mess. I cannot tell what you think the final answer should be. Please take the trouble to type out at least your final answer.

 

[vela]

Homework Statement

integration with respect to x

Homework Equations

integral 1/sqrt (a^2 - x^2) = arcsin(x/a)

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))

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

 

[vela]

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.

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

 

[Delta2]

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

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 $x\neq 0$

 

[Vriska]

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

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

 

[Vriska]

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?

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

 

[ehild]

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

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

 

[Vriska]

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

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?

 

[ehild]

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?

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

 

[PeroK]

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

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.

 

[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?

 

[stevendaryl]

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)}

 

[QuantumQuest]

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

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.

 

[Vriska]

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

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?

 

[ehild]

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?

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