「例山求f 2 解]令c-2=t2 2t 即x=ln(t2+2 t2+2 12t t=2 t tt2+2 +2 2· arctan+C √2 2 =√2 arctan 2021/2/20 2+C 6
2021/2/20 6 − = dx e I x 2 1 [例1] 求 [解] 2 e 2 t x 令 − = ln( 2), 2 即 x = t + dt t t dx 2 2 2 + = + = dt t t t I 2 1 2 2 + = dt t 2 1 2 2 C t = + 2 arctan 2 1 2 C e x + − = 2 2 2 arctan
例2求I=「√4-x2x 元 解」令x=2sint( <t<) V4-x=2v1-sinft= 2 cos t=2 cost =2 cost dx=2 cos tdt 1+cos 2t cos tdt =4 2 2(t+-sin 2t)+C 2021/2/20
2021/2/20 7 I = − x dx 2 [例2] 求 4 [ 解 ] 令 x = 2sin t ) 2 2 ( − t t x t t t 2cos 4 2 1 sin 2 cos 2cos 2 2 2 = − = − = = I = tdt 2 4 cos + = dt t 2 1 cos 2 4 = t + sin2t) + C 21 2( dx = 2costdt
为了作变量回代将改写为 1=2(t+sint cost)+C 根据代换函数x=2sint,作一个 直角三角形 4 2 =2 arcsin-+√4-x2+C 2021/2/20 8
2021/2/20 8 为了作变量回代,将I改写为 I = 2(t + sint cost) + C 直角三角形 根据代换函数x = 2sint, 作一个 x 2 2 4 − x t I = − x dx 2 4 x C x x = + − + 2 4 2 2 2arcsin
「例3求I= x2+9 解」令x=3tant x +9=3vtan t+1=3vsect=3sect d x= 3sec tdt 3sec t dt= sect dt x2+9 sect Insect +tan t+C 2021/2/20
2021/2/20 9 + = 9 [ 3 ] 2 x dx 例 求 I [ 解 ] 令 x = 3tan t x 9 3 tan t 1 3 sec t 3sec t 2 2 2 + = + = = = ln sec t + tan t + C = = + dt t dt tt x dx sec 3sec 3sec 9 2 2 dx tdt 2 = 3sec
Ⅰ= In/sect+tant+C x+ sect= v+9 3 3 dx 2 9 x +-+c x2+9 3 3 x+√x2+9 n +C=Inlx+vx+9+c 3 2021/2/20
2021/2/20 10 x t 3 9 2 x + 3 9 sec 2 + = x t 1 2 2 3 3 9 ln 9 c x x x dx + + + = + 1 2 3 9 ln c x x + + + = = ln x + x + 9 + c 2 I = lnsect + tant +C