my general concept question is in regards to the conic sections topic in calculus. I see that p is the height from the vertex to the focus of a horizontal parabola. When I am doing calculus and p is involved, do I treat p and it's exponents, and coefficients, as constants that do not change on integration or differentiation?
Thank you for any replies in advance.
$\left(\frac{1}{p}\right)$ $\int_0^{2p} \left(\sqrt {1 + \left(\frac{x}{2p}\right)^2}\right)$ which through trig substitution becomes
4p$\int_0^{2p} sec(\theta)^3$dx
The question is with respect to this integral. I am confused whether, or not to involve the 4p when I integrate by parts or to leave it out until after the integration is done.
Well, first of all your step is wrong. The two integrals aren't equal. In this answer I will help you evalute the integral.
We have:
$$\mathscr{I}_{\space\text{n}}:=\int_0^\text{n}\sqrt{1+\left(\frac{x}{\text{n}}\right)^2}\space\text{d}x\tag1$$
Substitute $\text{u}:=\arctan\left(\frac{x}{\text{n}}\right)$:
$$\mathscr{I}_{\space\text{n}}=\text{n}\cdot\int_{\arctan\left(\frac{0}{\text{n}}\right)}^{\arctan\left(\frac{\text{n}}{\text{n}}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}=\text{n}\cdot\int_0^\frac{\pi}{4}\sec^3\left(\text{u}\right)\space\text{d}\text{u}\tag2$$
Now, applying this (the proof is in the Wiki page), we can write:
$$\mathscr{I}_{\space\text{n}}=\text{n}\cdot\left[\frac{\sec\left(\text{u}\right)\cdot\tan\left(\text{u}\right)}{2}+\frac{\ln\left|\sec\left(\text{u}\right)+\tan\left(\text{u}\right)\right|}{2}\right]_0^\frac{\pi}{4}=\text{n}\cdot\frac{\sqrt{2}+\text{arcsinh}\left(1\right)}{2}\tag3$$