Question

Find the length of the curve Y=integral from 0 to X of sqrt(cos2t)dt from x=0 to x=pi/4. I'm confused on there being limits of integration within the Y when I'm supposed plug Y into the Arc Length function and use the limits of integration in that.

Answer 1

\[y=\int_0^x\sqrt{\cos 2t}~dt~~?\]

Answer 2

You don't actually need to evaluate the integral. Remember that the arc length integral formula uses the derivative of the curve:
\[L=\int_CdS=\int_a^b\sqrt{1+\left[\frac{dy}{dx}\right]^2}~dx\]
Find the derivative using the fundamental theorem of calculus,
\[\frac{d}{dx}\left[\int_c^{g(x)}f(t)~dt\right]=f(g(x))\cdot g'(x)\]
You're given the limits for the integral, \(\left[0,\dfrac{\pi}{4}\right]\), so you need to evaluate this integral:
\[\large L=\int_0^{\pi/4}\sqrt{1+\left(\sqrt{\cos2x}\right)^2}~dx=\int_0^{\pi/4}\sqrt{1+\cos2x}~dx\]

Answer 3

Identity to simplify the integral:\[1+\cos2x=1+2\cos^2x-1=2\cos^2x\]
\[\large L=\sqrt2\int_0^{\pi/4}\sqrt{\cos^2x}~dx=\sqrt2\int_0^{\pi/4}\cos x~dx\]