求定积分计算题及答案24542342222222442222lnsincosd,(sincos)d(sincos)d10,(cos)d0,(cos1.)d0.xIxxxJxxxKxxxxxIJxxKxxKIJ设,,比较三者的大小关系解析:由奇偶对称性可知,故2311112.()d,()e()d,()d(1)xfxxfxxfxxfxxxx设收敛且满足则2223131113111d()e,,(1)1dedd111ed,dln2,e(1)1A=(1ln2)exxxfxxAfxxAxxfxxxxAxxxxxxxx解析:令则对其两边积分得于是可得+2121111221121arctanxdx=_______xarctan111darctandarctandarctan11d441111[lnln(1)]ln2.23.424xxxxxxxxxxdxxxxxxxx求反常积分解析∣:212122211122221111212cosd_______112cos211dd2d11112(11)d4.xxxxxxxxxxxxxxxxxxx4.解析:330226202372320cos__________sin2d2sin3cos(sin5.)d326sin(1sin)d105axatxyatVyxatatttattta设星形线,则它绕轴旋转一周形成的旋转体体析:积为解1222222222111222220006s12limsin2sinsin__________.12limsin2sinin1limsin1sindsind()si.1ndnnnninnnnnnnnnnnniinnnxxxxxxxtttt解析:101000111007.d,d.d11dd()()d,d1dd.1dddnnxnxxnnnxxxnnnnnnnnnnnnnnnfxtfxttxtfxttfxtxtxtufuunntfxttfuufxnxxfxxnxn设连续则解析:得02000504401sin2d1sin2dcossind|cossin|d2sind442sin2sin228.4||()|()||xxxxxxxxxxxxxtdtttdt解析:211222221119.deedede1e1darctaneeeeeeeee244e.|()xxxxxxxxxxxx解析:222.10dlndd(ln)11ln.lnln|eeeexxxxxxxxx解析:002200d11.4dd()122arctan.222(4)2()xxxxxxxxx解析:∣233433333341133341040221m11lim(12)...