Question

Неопределенный интеграл

Answer 1

$\int \frac{dx}{3ctgx+2sinx}=[\, t=tg\frac{x}{2},\; x=2arctgt,\; dx=\frac{2dt}{1+t^2},\; sinx=\frac{2t}{1+t^2},\\\\cosx=\frac{1-t^2}{1+t^2},\; ctgx=\frac{cosx}{sinx}=\frac{1-t^2}{2t}\; ]=\int \frac{\frac{2dt}{1+t^2}}{\frac{3(1-t^2)}{2t}+\frac{2\cdot 2t}{1+t^2}}=\\\\=\int \frac{\frac{2\, dt}{1+t^2}}{\frac{(3-3t^2)(1+t^2)+4t\cdot 2t}{2t(1+t^2)}}=\int \frac{4t\, dt}{3+3t^2-3t^2-3t^4+8t^2}=\int \frac{4t\, dt}{-3t^4+8t^2+3}=\\\\=-\frac{4}{3}\int \frac{t\, dt}{t^4-\frac{8}{3}t^2-1}=-\frac{4}{3}\int \frac{t\, }{(t^2-\frac{8}{3\cdot 2})^2-\frac{16}{9}-1}=-\frac{4}{3}\int \frac{t\, dt}{(t^2-\frac{4}{3})^2-\frac{25}{9}}=$

$[\, z=t^2-\frac{4}{3}\; ,\; dz=2t\, dt\, ]=-\frac{4}{3\cdot 2}\int \frac{dz}{z^2-\frac{25}{9}}=\\\\=-\frac{2}{3}\cdot \frac{1}{2\cdot (5/3)}\cdot ln\Big |\frac{z-\frac{5}{3}}{z+\frac{5}{3}}\Big |+C=-\frac{1}{5}\cdot ln\Big |\frac{t^2-\frac{4}{3}-\frac{5}{3}}{t^2-\frac{4}{3}+\frac{5}{3}}\Big |+C=\\\\=-\frac{1}{5}\cdot ln\Big |\frac{tg^2\frac{x}{2}-3}{tg^2\frac{x}{2}+\frac{1}{3}}\Big |+C=-\frac{1}{5}\cdot ln\Big |\frac{3tg^2\frac{x}{2}-9}{3tg^2\frac{x}{2}+1}\Big |+C$