Question

решить определенный интеграл

Answer 1

$displaystyleintlimits^{2arctgfrac{1}{2}}_0frac{1+sinx}{(1-sinx)^2}dx=2intlimits^frac{1}{2}_0frac{(t+1)^2}{1+t^2}frac{(1+t^2)^2}{(t-1)^4}frac{dt}{1+t^2}=\=2intlimits^frac{1}{2}_0frac{(t+1)^2}{(t-1)^4}dt=2intlimits^frac{1}{2}_0(frac{1}{(t-1)^2}+frac{4}{(t-1)^3}+frac{4}{(t-1)^4})dt=\=2(-frac{1}{t-1}-frac{2}{(t-1)^2}-frac{4}{3(t-1)^3})|^frac{1}{2}_0=2(2-8+10frac{2}{3}-1+2-1frac{1}{3})=\=8frac{2}{3}$

$displaystyle t=tgfrac{x}{2};x=2arctgt;dx=frac{2dt}{1+t^2}\t_1=tg(arctgfrac{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$