Question

find the length of the curve $y = \int_{0}^{x} \sqrt{\cos 2t}dt$ from $x = 0$ to $x = \frac{\pi}{4}$. set up the integral to find the length of the curve. $l=square$ (type an exact answer, using $\pi$ as needed.)

Explanation:

Step1: Recall arc - length formula

The arc - length formula for a curve $y = F(x)=\int_{a}^{x}f(t)dt$ from $x = a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. First, find the derivative of $y$ using the Fundamental Theorem of Calculus. If $y=\int_{0}^{x}\sqrt{\cos2t}dt$, then by the Fundamental Theorem of Calculus, $y'=\sqrt{\cos2x}$.

Step2: Substitute $y'$ into arc - length formula

Substitute $y'=\sqrt{\cos2x}$ into the arc - length formula $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $a = 0$, $b=\frac{\pi}{4}$, so $L=\int_{0}^{\frac{\pi}{4}}\sqrt{1 + (\sqrt{\cos2x})^{2}}dx=\int_{0}^{\frac{\pi}{4}}\sqrt{1+\cos2x}dx$.

Step3: Use the double - angle identity

Recall the double - angle identity $\cos2x=2\cos^{2}x - 1$, then $1+\cos2x = 2\cos^{2}x$. So, $\sqrt{1+\cos2x}=\sqrt{2\cos^{2}x}=\sqrt{2}|\cos x|$. Since $x\in[0,\frac{\pi}{4}]$ and $\cos x\geq0$ on this interval, $\sqrt{1+\cos2x}=\sqrt{2}\cos x$.

Step4: Evaluate the integral

Now, evaluate the integral $L=\int_{0}^{\frac{\pi}{4}}\sqrt{2}\cos xdx$. Using the integral formula $\int\cos xdx=\sin x + C$, we have $L=\sqrt{2}[\sin x]_{0}^{\frac{\pi}{4}}=\sqrt{2}(\sin\frac{\pi}{4}-\sin0)$. Since $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and $\sin0 = 0$, $L=\sqrt{2}\times\frac{\sqrt{2}}{2}=1$.

Answer:

$1$