- #1
Odious Suspect
- 43
- 0
This arises in Joos's discussion of planetary motion, at the following URL:
https://books.google.com/books?id=wFl2MkpcY6kC&lpg=PP1&pg=PA91#v=onepage&q&f=false
I've modified the notation in obvious ways.
He asserts that the following expression is "the familiar arc-cosine form":
$$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$$
It's not familiar to me, and I am having trouble confirming it. The following is the approach I have taken.
$$\theta =\arccos (u)$$
$$\cos (\theta )=u$$
$$-\sin (\theta ) \frac{d\theta }{dx}=\frac{du}{dx}$$
$$\frac{d\theta }{dx}=-\frac{\frac{du}{dx}}{\sin (\theta )}=-\frac{\frac{du}{dx}}{\sqrt{1-u^2}}$$
$$u=\frac{b-hx}{\sqrt{a+b^2h}}$$
$$\frac{du}{dx}=-h$$
$$\frac{d\theta }{h}=\frac{dx}{\sqrt{1-u^2}}=\frac{d\arccos (u)}{h}$$
I seem to lose a minus sign that appears in the original form, but that may not be significant.
$$\int \frac{1}{\sqrt{1-u^2}} \, dx=\frac{\arccos (u)}{h}$$
$$\int \frac{1}{\sqrt{1-u^2}} \, dx=\frac{\arccos (u)}{h}$$
$$\int \frac{1}{\frac{\sqrt{1-u^2}}{\sqrt{h}}} \, dx=\frac{\arccos (u)}{\sqrt{h}}$$
$$\frac{1-u^2}{h}=\frac{1}{h}\left(1-\frac{b^2-2 bhh+x^2x^2}{a+b^2h}\right)=\frac{a+b^2h-\left(b^2-2 bhh+x^2x^2\right)}{h\left(a+b^2h\right)}=\frac{a2 b+hhx-h^2x^2}{h\left(a+b^2h\right)}=\frac{a+2 bx-hx^2}{a+b^2h}$$
That's where I get stuck. Notice that the numerator in the final expression is identical to the radicand of the denominator of the lhs in the first equation. In order for that to work, ##a+b^2h=1## must hold. No matter how I try, I just can't seem to pound it into submission.
$$C+\theta =-\int \frac{2 c}{\sqrt{\left(v_0^2-\frac{2 GM}{r_0}\right)+2 GM\mu -4 c^2 \mu ^2}} \, d\mu$$
$$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$$
$$x=\mu ;\text{Null}h=4 c^2;\text{Null}a=v_0^2-\frac{2 GM}{r_0};b=GM$$
$$a+b^2h=G^2 M^2+\left(V_0^2-\frac{2 GM}{r_0}\right)4c^2$$
How might I approach this problem effectively?
https://books.google.com/books?id=wFl2MkpcY6kC&lpg=PP1&pg=PA91#v=onepage&q&f=false
I've modified the notation in obvious ways.
He asserts that the following expression is "the familiar arc-cosine form":
$$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$$
It's not familiar to me, and I am having trouble confirming it. The following is the approach I have taken.
$$\theta =\arccos (u)$$
$$\cos (\theta )=u$$
$$-\sin (\theta ) \frac{d\theta }{dx}=\frac{du}{dx}$$
$$\frac{d\theta }{dx}=-\frac{\frac{du}{dx}}{\sin (\theta )}=-\frac{\frac{du}{dx}}{\sqrt{1-u^2}}$$
$$u=\frac{b-hx}{\sqrt{a+b^2h}}$$
$$\frac{du}{dx}=-h$$
$$\frac{d\theta }{h}=\frac{dx}{\sqrt{1-u^2}}=\frac{d\arccos (u)}{h}$$
I seem to lose a minus sign that appears in the original form, but that may not be significant.
$$\int \frac{1}{\sqrt{1-u^2}} \, dx=\frac{\arccos (u)}{h}$$
$$\int \frac{1}{\sqrt{1-u^2}} \, dx=\frac{\arccos (u)}{h}$$
$$\int \frac{1}{\frac{\sqrt{1-u^2}}{\sqrt{h}}} \, dx=\frac{\arccos (u)}{\sqrt{h}}$$
$$\frac{1-u^2}{h}=\frac{1}{h}\left(1-\frac{b^2-2 bhh+x^2x^2}{a+b^2h}\right)=\frac{a+b^2h-\left(b^2-2 bhh+x^2x^2\right)}{h\left(a+b^2h\right)}=\frac{a2 b+hhx-h^2x^2}{h\left(a+b^2h\right)}=\frac{a+2 bx-hx^2}{a+b^2h}$$
That's where I get stuck. Notice that the numerator in the final expression is identical to the radicand of the denominator of the lhs in the first equation. In order for that to work, ##a+b^2h=1## must hold. No matter how I try, I just can't seem to pound it into submission.
$$C+\theta =-\int \frac{2 c}{\sqrt{\left(v_0^2-\frac{2 GM}{r_0}\right)+2 GM\mu -4 c^2 \mu ^2}} \, d\mu$$
$$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$$
$$x=\mu ;\text{Null}h=4 c^2;\text{Null}a=v_0^2-\frac{2 GM}{r_0};b=GM$$
$$a+b^2h=G^2 M^2+\left(V_0^2-\frac{2 GM}{r_0}\right)4c^2$$
How might I approach this problem effectively?