I'm trying to calculate the following integral:
$\int_{0}^{\pi/3} \sqrt{\sec^2 (x)} dx$
But I have no idea where to start. Can you give me some advice?
$\endgroup$ 33 Answers
$\begingroup$$$=\int_{0}^{\frac{\pi}{3}}\sec{x}dx$$$$=\int_{0}^{\frac{\pi}{3}}\frac{\sec{x}(\sec{x}+\tan{x})}{\sec{x}+\tan{x}}dx$$$$=\int_{0}^{\frac{\pi}{3}}\frac{\sec^2{x}+\sec{x}\tan{x}}{\tan{x}+\sec{x}}dx$$$$=\Big[\log{(\sec{x}+\tan{x})}\Big]_0^{\frac{\pi}{3}}$$$$=\log{(2+\sqrt{3})}$$
$\endgroup$ $\begingroup$As on the integration interval, $\sec x>0$, we may simplify to$$\int_0^\tfrac\pi3\sec x\,\mathrm dx=\int_0^\tfrac\pi3\frac{\mathrm dx}{\cos x}=\int_0^\tfrac\pi3\frac{\cos x\,\mathrm dx}{\cos^2 x}.$$Can you continue?
$\endgroup$ $\begingroup$Find the interval(s) where $\sec(x)$ is positive vs. negative in $[0,\pi/3].$ then you can get rid of the absolute values (inserting $-$ if negative) and integrate $\sec(x)$ instead.
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