Find the length of the curve.

Question

Let $x(t) = \sin(t)^4$, $y(t) = \cos(t)^2$, $t \in [0, \frac{\pi}{2}]$

We want to find this curve's length.

There lots of these type questions, but I have a problem with just counting an answer :

Here's what I know: $l = \int_0^{\frac{\pi}{2}}\sqrt{x'(t)^2 + y'(t)^2}{\rm d}t$

$x'(t) = 4\sin^3(t)\cos(t), y'(t) = -2\cos(t)\sin(t)$

We substitute the resulting into the formula: $l = \int_0^{\frac{\pi}{2}}\sqrt{16\sin^6(t)\cos^2(t) + 4\cos^2(t)\sin^2(t)}{\rm d}t = $

$= \int_0^{\frac{\pi}{2}}2\cos(t)\sin(t)\sqrt{4\sin^4(t) + 1}{\rm d}t$

But what can I do next to get an answer ? I don't know any technique for this. Can someone help me ?

Answer 1

Put $s=4\sin^{2}t+1$. You get $\frac 1 4 \int_1^{5}\sqrt {s}ds$.

EDIT. It should have been $\sqrt {4\sin^{4}t+1}$ inside the integral instead of $\sqrt {4\sin^{2}t+1}$. So you have to make the substitution $s=4\sin^{4}t+1$ and use the fact that $\sin^{2}t= \sqrt {\frac {s-1} 4}$.

Answer 2

HINT: Substitute $u = 4\sin^2(x) + 1$.