Question

Кто может помочь решить интеграл

Answer 1

$\displaystyle\int\frac{sinxdx}{sin^2x-3sinx+2}=\int\frac{2tdt}{(t-1)^2(t^2-t+1)}=\\=2\int(\frac{d(t-1)}{(t-1)^2}-\frac{d(t-\frac{1}{2})}{(t-\frac{1}{2})+\frac{3}{4}})dt=-\frac{2}{t-1}-\sqrt3arctg\frac{2t-1}{\sqrt3}+C=\\=-\frac{2}{tg\frac{x}{2}-1}-\sqrt3arctg\frac{2tg\frac{x}{2}-1}{\sqrt3}+C$
$\displaystyle t=tg\frac{x}{2};x=2arctgt;dx=\frac{2dt}{1+t^2}\\sinx=\frac{2t}{1+t^2}\\sin^2x-3sinx+2=\frac{4t^2}{(1+t^2)^2}-\frac{6t}{1+t^2}+2=\\=\frac{(2t^4-6t^3+8t^2-6t+2)}{(1+t^2)^2}=\frac{2(t-1)^2(t^2-t+1)}{(1+t^2)^2}\\sinxdx=\frac{4tdt}{(1+t^2)^2}\\\frac{2t}{(t-1)^2(t^2-t+1)}=\frac{A}{(t-1)}+\frac{B}{(t-1)^2}+\frac{Ct+D}{t^2-t+1}=\\=\frac{2}{(t-1)^2}-\frac{2}{t^2-t+1}\\2t=A(t^3-2t^2+2t-1)+B(t^2-t+1)+(Ct+D)(t^2-2t+1)\\t^3|0=A+C\\t^2|0=-2A+B-2C+D\\t|2=2A-B+C-2D\\t^0|0=-A+B+D\\A=0;B=2;C=0;D=-2$