# calculating the arc length of a function

The function is: $$f(x) =2\ln(4-x^2)$$ and the length of the arc is from $$-1$$ to $$1$$.

So i know the formula for calculating the arc length is$$\int_{a}^{b}\sqrt{1+(f'(x))^2} dx$$ the derivative of the function $$f(x)$$ is, if i am correct $$2\frac{d}{dg}\ln(g)\frac{d}{dx}(4-x^2)$$ with $$g = (4-x^2)$$. So$$f'(x)= -\cfrac{4x}{4-x^2}$$and $$(f'(x))^2=\cfrac{16x^2}{(4-x^2)^2}$$ and the formula becomes: $$\int_{-1}^{1}\sqrt{1+\cfrac{16x^2}{(4-x^2)^2}}dx$$ the problem is when i use the $$u$$ substitution of $$u=1+\cfrac{16x^2}{(4-x^2)^2}$$ And i want to make new boundaries for the integration $$i$$ have $$2$$ the same numbers (both are 2,77). I don't know what i have done wrong but the answer must be $$4\ln(3)-2$$

## 3 Answers

The substitution you're suggesting won't help simplify your problem.

$$1+\frac{16x^2}{(4-x^2)^2}$$

$$=\frac{(4-x^2)^2 + 16x^2}{(4-x^2)^2}$$ $$=\frac{(4+x^2)^2}{(4-x^2)^2}$$

This will help solve.

To the question you have posed, this is happening because the function is even. And if $$f(x)$$ is an even function, then:

$$\int_{-1}^{1} f(x) dx = 2 \int_{0}^{1} f(x) dx$$

It turns out that$$(\forall x\in[-1,1]):\sqrt{1+f'(x)^2}=\frac{4+x^2}{4-x^2}=\frac2{x+2}-\frac2{x-2}-1$$and my guess is that you can take it from here.

When you do a $$u$$ substitution, you have to make sure that $$u$$ is an injective function over the limits of integration. The way you've defined your $$u$$, however, $$u(-x)=u(x)$$, so it's not injective. When you substitute $$du$$ in for $$dx$$, you should find that $$dx = \frac {du}{h'(x)}$$, where $$h(x) = 1+\cfrac{16x^2}{(4-x^2)^2}$$. But at $$x=0$$, $$h'(0)=0$$, making this an improper integral. You can fix this by splitting the integral into an integral from $$-1$$ to $$0$$ (strictly speaking, to be rigorous you have to take the integral from $$-1$$ to $$-\epsilon$$ and take the limit as $$\epsilon$$ goes to zero, but you can get away with fudging on that point), and another integral from $$0$$ to $$1$$. $$u$$ will then be injective within each integral. And since $$u$$ is symmetric, you can simply take one integral and double the answer you get.