4. Find the length $s$ of the curve $x(t) = \tan t$, $y(t) = \frac{1}{2}(\sec^2 t + 1)$ on the given interval $0 \le t \le \frac{\pi}{4}$. (5 points)
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Then $x'(t) = \sec^2 t$ and $y'(t) = \sec^2 t \tan t$. The arc length is given by $$ s = \int_0^{\pi/4} \sqrt{(x'(t))^2 + (y'(t))^2} dt = \int_0^{\pi/4} \sqrt{(\sec^2 t)^2 + (\sec^2 t \tan t)^2} dt $$ $$ = \int_0^{\pi/4} \sqrt{\sec^4 t + \sec^4 t \tan^2 t} dt = Show more…
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