例14求∫ cos xdx 2 解∫ cos xdx 1+ cos 2x +2cos 2x+cos 2x 1+cos 4x =1+2c0s2x+ dx 2 cos 4x +2c0s2x+ 4叭2 2 x+ sine H sin4x +C 3 =-x+-sin 2x+-sin 4x +C 32
cos . 4 求 xdx sin4 , 4 1 2 1 sin2 2 3 4 1 x x x + C = + + xdx 4 cos + = x x d 2 2 1 cos 2 ( ) = 1+ 2cos2x + cos 2x dx 4 1 2 + = + + x x x d 2 1 cos4 1 2cos2 4 1 = + + x x x d 2 cos4 2cos 2 2 3 4 1 sin4 . 32 1 sin2 4 1 8 3 = x + x + x + C 例14 解
例15求 scram 解∫ scram dx Snd 2sin --cos d tan tan -cos tan 2 2 In tan +C ∵∴tan cos x cScr-cotr, SInd Iescxdx= Inescx-cotx+C
csc . 求 xdx csc xdx = x x d sin 1 = x x x d 2 cos 2 2sin 1 = 2 2 cos 2 tan 1 2 x x x d = 2 tan 2 tan 1 x x d . 2 ln tan C x = + 2 tan x x x sin 1− cos = = csc x − cot x, csc xdx= ln csc x − cot x + C. 例15 解