Question

Кто может помочь решить интеграл

Answer 1

$displaystyleintfrac{sinxdx}{sin^2x-3sinx+2}=intfrac{2tdt}{(t-1)^2(t^2-t+1)}=\=2int(frac{d(t-1)}{(t-1)^2}-frac{d(t-frac{1}{2})}{(t-frac{1}{2})+frac{3}{4}})dt=-frac{2}{t-1}-sqrt3arctgfrac{2t-1}{sqrt3}+C=\=-frac{2}{tgfrac{x}{2}-1}-sqrt3arctgfrac{2tgfrac{x}{2}-1}{sqrt3}+C$
$displaystyle t=tgfrac{x}{2};x=2arctgt;dx=frac{2dt}{1+t^2}\sinx=frac{2t}{1+t^2}\sin^2x-3sinx+2=frac{4t^2}{(1+t^2)^2}-frac{6t}{1+t^2}+2=\=frac{(2t^4-6t^3+8t^2-6t+2)}{(1+t^2)^2}=frac{2(t-1)^2(t^2-t+1)}{(1+t^2)^2}\sinxdx=frac{4tdt}{(1+t^2)^2}\frac{2t}{(t-1)^2(t^2-t+1)}=frac{A}{(t-1)}+frac{B}{(t-1)^2}+frac{Ct+D}{t^2-t+1}=\=frac{2}{(t-1)^2}-frac{2}{t^2-t+1}\2t=A(t^3-2t^2+2t-1)+B(t^2-t+1)+(Ct+D)(t^2-2t+1)\t^3|0=A+C\t^2|0=-2A+B-2C+D\t|2=2A-B+C-2D\t^0|0=-A+B+D\A=0;B=2;C=0;D=-2$