Question

Необходимо решить интеграл: tg(x)/(1-ctg^2(x))

Answer 1

$\int \frac{tgx\, dx}{1-ctg^2x}=[\; t=tgx\; ,\; x=arctgt\; ,\; dx=\frac{dt}{1+t^2}\; ]=\int \frac{t\, dt}{(1-\frac{1}{t^2})(1+t^2)}=\\\\=\int \frac{t^3\, dt}{(t^2-1)(1+t^2)}=Q\\\\\frac{t^3}{(t-1)(t+1)(t^2+1)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{Ct+D}{t^2+1}\\\\t^3=A(t+1)(t^2+1)+B(t-1)(t^2+1)+(Ct+D)(t-1)(t+1)\\\\t=1:\; \; A=\frac{t^3}{(t+1)(t^2+1)}=\frac{1}{2\cdot 2}=\frac{1}{4}\\\\t=-1:\; \; B=\frac{t^3}{(t-1)(t^2+1)}=-\frac{1}{4}\\\\\\t^3\; |\; A+B+C=1\; ,\; \; C=1-A-B=1-\frac{1}{4}+\frac{1}{4}=1$

$t^2\; |\; A-B+D=0\; ,\; \; D=-A+B=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}\\\\\\Q=\frac{1}{4}\int \frac{dt}{t-1}-\frac{1}{4}\int \frac{dt}{t+1}+\int \frac{t-\frac{1}{2}}{t^2+1}\, dt=\frac{1}{4}\cdot ln|t-1|-\frac{1}{4}\cdot ln|t+1|+\\\\+\frac{1}{2}\int \frac{2t\, dt}{t^2+1}-\frac{1}{2}\int \frac{dt}{t^2+1}=\frac{1}{4}\cdot ln\Big |\frac{t-1}{t+1}\Big |+\frac{1}{2}\cdot ln|t^2+1|-\frac{1}{2}\cdot arctgt+C=\\\\=\frac{1}{4}\cdot ln\Big |\frac{tgx-1}{tgx+1}\Big |+\frac{1}{2}\cdot \Big (ln|tg^2x+1|-artg\, (tgx)\Big)+C=$

$=\frac{1}{4}\cdot ln\Big |\frac{tgx-1}{tgx+1}\Big |+\frac{1}{2}\cdot \Big (ln(tg^2x+1)-x\Big)+C=$

$=\frac{1}{4}\cdot ln\Big |\frac{tgx-1}{tgx+1}\Big |+ln\sqrt{tg^2x+1}-\frac{1}{2}x+C$