Question

решить определенный интеграл

Answer 1

$\displaystyle\int\limits^{2arctg\frac{1}{2}}_0\frac{1+sinx}{(1-sinx)^2}dx=2\int\limits^\frac{1}{2}_0\frac{(t+1)^2}{1+t^2}\frac{(1+t^2)^2}{(t-1)^4}\frac{dt}{1+t^2}=\\=2\int\limits^\frac{1}{2}_0\frac{(t+1)^2}{(t-1)^4}dt=2\int\limits^\frac{1}{2}_0(\frac{1}{(t-1)^2}+\frac{4}{(t-1)^3}+\frac{4}{(t-1)^4})dt=\\=2(-\frac{1}{t-1}-\frac{2}{(t-1)^2}-\frac{4}{3(t-1)^3})|^\frac{1}{2}_0=2(2-8+10\frac{2}{3}-1+2-1\frac{1}{3})=\\=8\frac{2}{3}$

$\displaystyle t=tg\frac{x}{2};x=2arctgt;dx=\frac{2dt}{1+t^2}\\t_1=tg(arctg\frac{1}{2})=\frac{1}{2};t_2=tg0=0\\1+sinx=1+\frac{2t}{1+t^2}=\frac{t^2+2t+1}{1+t^2}=\frac{(t+1)^2}{1+t^2}\\(1-sinx)^2=(1-\frac{2t}{1+t^2})^2=(\frac{t^2-2t+1}{1+t^2})^2=\frac{(t-1)^4}{(1+t^2)^2}\\\\\frac{(t+1)^2}{(t-1)^4}=\frac{A}{t-1}+\frac{B}{(t-1)^2}+\frac{C}{(t-1)^3}+\frac{D}{(t-1)^4}\\t^2+2t+1=A(t^3-3t^2+3t-1)+B(t^2-2t+1)+C(t-1)+D\\t^3|0=A\\t^2|1=-3A+B\\t|2=3A-2B+C\\t^0|1=-A+B-C+D\\A=0;B=1;C=4;D=4$