# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Preliminary Questions - Page 468: 1

$$\int_0^{\pi/4}\sqrt{1+\sin^2x}dx.$$

#### Work Step by Step

To find the length of the curve $y=\cos x$, we fist calculate $y'=-\sin x$. Then we have the integral, $$\int_0^{\pi/4}\sqrt{1+(y')^2}dx=\int_0^{\pi/4}\sqrt{1+\sin^2x}dx.$$ So the right answer is $$\int_0^{\pi/4}\sqrt{1+\sin^2x}dx.$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.