例5求 tan xdx 解 tanxi= sIn cos d cosx cosr In cosx+C 请问:「 cot xdx=0sxa= Inin x+C; sInd Economic-mathematics 48-11 Wednesday, February 24, 2021
Economic-mathematics 48- 11 Wednesday, February 24, 2021 例5 求 解 tan . xdx = dx x x xdx cos sin tan = − x d x cos cos = 请问： cot xdx dx x x sin cos = lnsin x + C; = − lncos x + C;
例6求|cos2xax. 解∫ 2 1+cos2x cos d dx 2 订∫+1cos2xh x+「cos2x(2x) 24 +-sin 2x+C 2 类似可得∫snth=1 sin 2x + c: 24 Economic-mathematics 48-12 Wednesday, February 24, 2021
Economic-mathematics 48- 12 Wednesday, February 24, 2021 cos . 2 xdx sin 2 C; 4 1 2 sin2 = − + x x 类似可得 xdx + = dx x xdx 2 1 cos2 cos2 = dx + cos2xdx 2 1 2 1 = + cos2 (2 ) 4 1 2 xd x x 例6 求 解 sin 2 . 4 1 2 x C x = + +
例7求[ cscad 解1 esche=∫ = 1x2 dx SIn 2sin--cos 2 d()= d (tan tan(cos x22 tan In tan+C=In(esc x-cotx)+C Economic-mathematics 48-13 Wednesday, February 24, 2021
Economic-mathematics 48- 13 Wednesday, February 24, 2021 例7 求 解 1 = dx sin x 1 csc . xdx csc xdx = dx x x 2 cos 2 2sin 1 = ) 2 ( ) 2 (cos 2 tan 1 2 x d x x = ) 2 (tan 2 tan 1 x d x C x = + 2 ln tan = ln(csc x − cot x)+C
例7求[ cscad sInd dx sIn x 解2∫cexd=∫ d(cos x) SInx 1-cos-x 十 ) d(cos x) 2J1-cosx 1+cosx In(1-cos x)-In (1+cos x)1+C 1-cosx n +C=In(cscx-cot x)+C. 2 1+cosx 类似地可推出∫scdx=m(ex+tmx)+C Economic-mathematics 48-14 Wednesday, February 24, 2021
Economic-mathematics 48- 14 Wednesday, February 24, 2021 解 2 = dx sin x 1 csc xdx dx x x 2 sin sin − = − (cos ) 1 cos 1 2 d x x + + − = − ) (cos ) 1 cos 1 1 cos 1 ( 2 1 d x x x ln(csc cot ) . 1 cos 1 cos ln 2 1 C x x C x x + = − + + − = 类似地可推出 sec ln(sec tan ) . xdx = x + x + C 例7 求 csc . xdx = [ln(1− cos x) − ln(1+ cos x)]+ C 2 1
第二类换元积分法 问题x51-x2=?引进第二类换元积分法 第二换元法是适当选择=q(,将积分f(x)t 化为积分f(q(t)y'()lt 于是∫f(x)tk=f(o()p(t ja(odt=G(0)+C=G(-1(x)+C; 其中代换x=(t)应满足 (1)g'(t)连续 (2)g'(t)≠0,以保证反函数q(t)的存在 Economic-mathematics 48-15 Wednesday, February 24, 2021
Economic-mathematics 48- 15 Wednesday, February 24, 2021 (1) (t)连续. 化为积分 第二换元法是适当选择 将积分 x = (t), f (x)d x ( ( )) ( ) . f t t dt 于是 f (x)dx = f ((t))(t)dt ( ) ( ) ( ( )) ; 1 = g t dt = G t + C = G x + C − 其中代换x =(t)应满足 (2) ( ) 0, ( ) . t 以保证反函数 −1 t 的存在 问题 1 ? 5 2 − = x x dx 引进第二类换元积分法 二、第二类换元积分法