Question

Решить неопределённый интеграл

Answer 1

$\displaystyle \int\frac{dx}{4sinx+3cosx}=2\int\frac{dt}{-3t^2+8t+3}=2\int\frac{dt}{-(3t^2-8t+\frac{16}{3})+\frac{25}{3}}=\\=-\frac{2}{\sqrt3}\int\frac{d(\sqrt3t-\frac{4}{\sqrt{3}})}{(\sqrt3t-\frac{4}{\sqrt{3}})^2-\frac{25}{3}}=-\frac{1}{5}ln|\frac{t-3}{3t+1}|+C=\\=-\frac{1}{5}ln|\frac{tg\frac{x}{2}-3}{3tg\frac{x}{2}+1}|+C$
$t=tg\frac{x}{2};2arctgt=x\\dx=\frac{2dt}{1+t^2}\\sinx=\frac{2t}{1+t^2};cosx=\frac{1-t^2}{1+t^2}\\4sinx+3cosx=\frac{8t}{1+t^2}+\frac{3-3t^2}{1+t^2}=\frac{-3t^2+8t+3}{1+t^2}$