广州大学：《高等数学》课程教学资源（教案讲义）三角有理函数积分

§4三角有理函数积分 三角函数有理式」(smnx, cos xdx 型的积分 t=tg 万能代换:万能代换常用于三角函数有理式的积分,令 X 就有 COSx= tg dx= rctgf 1+cos x tg 解法一(用万能代换) d(-)=驾g 解法二(用初等化简) 解法三(用初等化简,并凑微) 1-cos x d sin x dx=csc xdx- 1-cos"x tgx+ tc=csc x -ctg+c=tg 代换法是一种很灵活的方法. 1+sin x dx 例1求mx(1+csx),(以下采用 Matlab帮助计算) f=2*(1+sin(x)/(sin(x)*(1+cos(x)*(1+t2))

§4 三角有理函数积分 三角函数有理式 型的积分 万能代换: 万能代换常用于三角函数有理式的积分， 令 ， 就有 ， ， . 解法一 ( 用万能代换 ) . 解法二 ( 用初等化简 ) . 解法三 ( 用初等化简, 并凑微 ) 代换法是一种很灵活的方法. 例 1 求 ， (以下采用 Matlab 帮助计算) f='2*(1+sin(x))/(sin(x)*(1+cos(x))*(1+t^2))';

flsimplify(subs(subs(f, 2*t/(1+t 2)', sin(x)), ' (1-t 2)/(1+t 2', 'cos(x) expand(f1) ans=1/2/t+1/2*t+1 1+sin x x(1+cos x) (f1) ans=1/4*t2+t+1/2*log(t) 1+si 2,+n1+c 例2求 f=2*(3-sin(x))/((3+cos(x))*(1+t^2)) f1= simplify(subs(subs(f,2*t/(1+t^2)’,sin(x)’),”(1-t^2)/(1+t^2)’,"’cos(x) f1=(3+3*t^2-2*t)/(2+t^2)/(1+t^2) 利用部分分式展开,积分 nt(f1) f2=1og(2+t^2)+3/2*2(1/2)*atan(/2*t*2^(1/2)-log(1+t^2) dx=In(2+t) 2 clgt-In(1+t)+C 例 求

f1=simplify(subs(subs(f,'2*t/(1+t^2)','sin(x)'),'(1-t^2)/(1+t^2)','cos(x) ')) f1 = 1/2*(1+t^2+2*t)/t expand(f1) ans = 1/2/t+1/2*t+1 int(f1) ans = 1/4*t^2+t+1/2*log(t) 例 2 求 f='2*(3-sin(x))/((3+cos(x))*(1+t^2))'; f1=simplify(subs(subs(f,'2*t/(1+t^2)','sin(x)'),'(1-t^2)/(1+t^2)','cos(x) ')) f1 = (3+3*t^2-2*t)/(2+t^2)/(1+t^2) 利用部分分式展开，积分 f2=int(f1) f2 =log(2+t^2)+3/2*2^(1/2)*atan(1/2*t*2^(1/2))-log(1+t^2) 例 3 求

dt 解 tg f=1/(a2*t^2+b^2)’;int(f) ans =1/b/a:atan(a*kt/b) dtg arc tg-tgx+C 例4 s='t-sart((x+2)/(x-2)): x=solve(f) X=2*(1+t^2)/(-1+t^2) simplify(diff (x, 't') ans=-8*t/(-1+t^2)^2 g=-8*t2/(x*(t2-1)2) gl=simplify(subs(g, x',2*(t 2+1)/(t 2-1),)) g1=-4*t2/(1+t^2)/(t2-1) ans =-log(t-1)+log(1+t)-2*atan(t) 1=2 t+1 dx=In 2arctgt +C - dx 1+x)√2+x (1+x)(1+x)(2-x) f=t2(2-x)/(1+x): x=solve(f), dx=simplify x=-(t^2-2)/(1+t^2) dx=-6*t/(1+t2)2 g=(-6/(1+x)2)/(1+t^2)2

解 f='1/(a^2*t^2+b^2)'; int(f) ans = 1/b/a*atan(a*t/b) 例 4 s='t-sqrt((x+2)/(x-2))';x=solve(f) x = 2*(1+t^2)/(-1+t^2) simplify(diff(x,'t')) ans = -8*t/(-1+t^2)^2 g='-8*t^2/(x*(t^2-1)^2)'； g1=simplify(subs(g,'x','2*(t^2+1)/(t^2-1)')) g1 = -4*t^2/(1+t^2)/(t^2-1) int(g1) ans = -log(t-1)+log(1+t)-2*atan(t) f='t^2-(2-x)/(1+x)';x=solve(f), dx=simplify x = -(t^2-2)/(1+t^2) dx = -6*t/(1+t^2)^2 g='(-6/(1+x)^2)/(1+t^2)^2';

1= simplify(subs(g,’x',’(2-t^2)/(1+t^2)’) 1=-2/3 (1+x)√2+ 1 -=dt d (x-1)2(x+2) +2)[x-1)f(x+2)]2 dx (x+2 f='y3-(x-1)/(x+2): x=solve(f), dx=simplify(diff(x, 'y)) (2*y3+1)/(y3-1) g=1/((x+2)*y2)*(9*y2/(y^3-1)-2) gl=simplify(subs(g, x','(2*y 3+1)/(1-y 3))) g1=-3/(y3-1) (g1) ans=-1og(y-1)+1/2*1og(y2+y+1)+3(1/2)*atan(1/3*(2*y+1)*3(1/2) In ly-1+In ly tg(2 3(x-12(x+2)

g1=simplify(subs(g,'x','(2-t^2)/(1+t^2)')) g1 = -2/3 f='y^3-(x-1)/(x+2)'; x=solve(f), dx=simplify(diff(x,'y')) x =-(2*y^3+1)/(y^3-1); dx = 9*y^2/(y^3-1)^2 g='1/((x+2)*y^2)*(9*y^2/(y^3-1)^2)'; g1=simplify(subs(g,'x','(2*y^3+1)/(1-y^3)')) g1 =-3/(y^3-1) int(g1) ans = -log(y-1)+1/2*log(y^2+y+1)+3^(1/2)*atan(1/3*(2*y+1)*3^(1/2))