Everyday Math 每日數學 之 Integration #20190529 (+solution)

Let $ u=\sqrt{x} $
Then $ x=u^2 $
$ dx=2udu $
$ \int \sec^2{\sqrt{x}}dx $
$ =\int 2u\sec^2 udu $
$ =\int 2ud(\tan u) $
$ =2u\tan u-2\int \tan u du $
$ =2u\tan u-2\int \frac{\sin u}{\cos u} du $
$ =2u\tan u+2\int \frac{1}{\cos u} d(\cos u) $
$ =2u\tan u+2\ln \left|\cos u \right| +C$
$ =2\sqrt{x}\tan \sqrt{x}+2\ln \left|\cos \sqrt{x} \right| +C$