⑩串紫学 Teaching Plan on Advanced Mathematics 例10求 dk (a>o x 解:令x= sect di= asecttan tdt t∈0, a sect. tan t dx= 2 x -a a tant =I sectat In(sect tant)+C r -l r -ll +cTianjin Polytechnic University Teaching Plan on Advanced Mathematics ( 0). 1 2 2  −  dx a x a 令 x = asect         2 dx = asecttantdt t 0, = −  dx x a 2 2 1 dt a t a t t   tan sec tan =  sectdt = ln(sec t + tant) +C t a x 2 2 x − a ln . 2 2 C a x a a x +        − = + 例10 求 解：