# Finding the length of a curve?

With the information given:

$$x=\frac{y^4}{8}+\frac{1}{4y^2}\,,\ \ 1 \le y \le 2$$

I must find the exact length of the curve.

I use this formula to find it: $$\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy$$

So of course, I should find what 1 + (dx/dy)^2 is.

This is what I got:

$$\frac{y^6 -y^{-6}+2}{4}$$

So . . .

$$\int_1^2\sqrt{\frac{y^6-y^{-6}+2}{4}} \, dy$$

I don't know... It just looks really funky to me. Should I do a $u$ substitution? Like $u=y^6-y^{-6}+2$?

There was a little mistake made when squaring, you should have $$\frac{y^6+2+y^{-6}}{4}\tag{1}$$ inside the square root. And the square root of (1) is very nice, in our interval it is $$\frac{y^3+y^{-3}}{2}.$$
• To integrate, no substitution. You are just integrating powers. For $y^3$ you get $\frac{1}{4}y^4$. For $y^{-3}$ you get $\frac{1}{-2}y^{-2}$. – André Nicolas Sep 28 '14 at 20:13
Observe that $$\frac{dx}{dy}=\frac{1}{2}\left(y^{3}-y^{-3}\right),$$ hence $$1+\left(\frac{dx}{dy}\right)^2=\frac{1}{2}\left(y^{3}+y^{-3}\right)^2$$ and thus $$\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=\int_{1}^{2} \frac{(y^3+y^{-3})\,dy}{2}.$$