#### Answer

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

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