4 三角有理函数积分

§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))';

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

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';

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))