Question

Интеграл и Предел по правилу Лопиталя. Помогите люди добрые. Только те кто понимает

Answer 1

$\int \dfrac{dx}{sinx+tgx}=\int \dfrac{dx}{sinx+\frac{sinx}{cosx}}=\int \dfrac{cosx\, dx}{sinx\cdot cosx+sinx}=\int \dfrac{cosx\, dx}{sinx\cdot (cosx+1)}=\\\\\\=\Big[\ t=tg\dfrac{x}{2}\ ,\ sinx=\dfrac{2t}{1+t^2}\ ,\ cosx=\dfrac{1-t^2}{1+t^2}\ ,\ dx=\dfrac{2\, dt}{1+t^2}\Big]=\\\\\\=\int \dfrac{\dfrac{1-t^2}{1+t^2}\cdot \dfrac{2\, dt}{1+t^2}}{\dfrac{2t}{1+t^2}\cdot \Big(\dfrac{1-t^2}{1+t^2}+1\Big)}=\int \dfrac{2(1-t^2)\, dt}{2t\cdot (1-t^2+1+t^2)}=\int \dfrac{(1-t^2)\, dt}{2t}=$

$=\int \Big(\dfrac{1}{2t}-\dfrac{t}{2}\Big)\, dt=\dfrac{1}{2}\int \dfrac{dt}{t}-\dfrac{1}{2}\int t\, dt=\dfrac{1}{2}\cdot ln|t|-\dfrac{1}{2}\cdot \dfrac{t^2}{2}+C=\\\\\\=\dfrac{1}{2}\cdot ln\Big|tg\dfrac{t}{2}\Big|-\dfrac{1}{4}\cdot tg^2\dfrac{x}{2}+C$