## Calculus (3rd Edition)

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