$y=1-x^2$, find length of curve in first quadrant
$\int ds= \int \sqrt{ (dx^2+dy^2)}=\int \sqrt{ (1+dy^2)}dx$
$\int_{0} ^1\sqrt{1+(-2x)^2}=\sqrt{1+4x^2}=\sqrt{4({\frac{1}{4}+x^2})}=2\sqrt{({\frac{1}{4}+x^2})}$
$x=\frac{1}{2}tan \theta$
$\frac{dx}{d\theta}=\frac{1}{2}cos \theta$
$\int_{0} ^{tan^-2}2\sqrt{({\frac{1}{4}+\frac{1}{4}tan^2\theta})}\frac{1}{2}sec^2 \theta d\theta$= $\frac{1}{2}sec ^3\theta d\theta$
is this right? and also when i changed to $\theta$, is the boundary right? i got very nice integral but the boundary is bad
Sorry Vik, I was being too careless. The problem is first that your are choosing $x = (1/2)\sin(\theta)$. But this way, $x$ has a max of only $1/2$, when we need it to reach all the way to $1$. Plus, your integration is wrong. You are trying to say that $1+\sin^2(x) = \cos^2(x)$, when the correct expression is $1-\sin^2(x) = \cos^2(x)$. I would recommend that you use the Pythagorean identity with $\sec$ and $\tan$ for the trig substitution. $$1+\tan^2(\theta) = \sec^2(\theta)$$.