§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))