<s>[INST] Approximating arc length I have the integral expression for the arc length: 
$$ \int \sqrt{1 + \frac{1}{4x}\,}\,\mathrm{d}x $$
and need to approximate the arc length of the curve: $y = $$ \sqrt{x}$
in between x = 0 and x = 4.
Should I be using the trapezoidal rule or Simpson's rule for this? And how can I begin to apply it?
 [/INST] The formula for arc length of $y = \sqrt{x}$ from $x = 0$ to $x=4$ is
$$ \int_{0}^{4} \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx$$
where here $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$, squaring this results in $\left( \frac{dy}{dx} \right)^2 = \frac{1}{4x}$. Hence our task is to integrate
$$\int_{0}^{4} \sqrt{1 + \frac{1}{4x}} dx$$
Let $$u = \sqrt{ 1 + \frac{1}{4x} } \Rightarrow u^2 = 1 + \frac{1}{4x} \Rightarrow 2udu = \frac{-1}{4x^2}dx$$
From that we also obtain that $x = \frac{1}{4(u^2 - 1)}$ so that $x^2 = \frac{1}{16(u^2 - 1)^2}$, then $\frac{-1}{4x^2} = -4(u^2-1)^2$, so that we get 
$$dx = \frac{-u}{2(u^2 -1)^2}du$$
Hence, with our sub, we get the integral
$$\int_{x=0}^{x=4} \frac{-u^2}{2(u^2 - 1)^2} du$$
Now I would do another sub, letting $u = \sec(\theta)$ so that $du = \sec(\theta)\tan(\theta)d\theta$, and we get with Phythagoras' Theorem that
$$\int_{x=0}^{x=4} \dfrac{-\sec^2(\theta)}{2\tan^4(\theta)} \sec(\theta)\tan(\theta)d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \dfrac{\sec^3(\theta)}{\tan^3(\theta)} d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \csc^3(\theta)d\theta,$$
This can be solved via integration by parts and is definitely done in detail somewhere on the internet. You should arrive at the answer
$$= \dfrac{-1}{2} \left( \dfrac{-\csc(\theta)\cot(\theta) + ln|\csc(\theta) - \cot(\theta)|}{2} \right)$$
I will leave it to you to do the back substitutions and plug in the bounds. Also, I feel like I pulled out a sledge hammer to solve this, I'm sure someone will post a more elegant solution.
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