Question

помогите пожалуйста с ответом

Answer 1

Ответ:

$\displaystyle \int \frac{du}{cosu}=\Big[\ t=tg\frac{u}{2}\ ,\ cosu=\frac{1-t^2}{1+t^2}\ ,\ du=\frac{2\, dt}{1+t^2}\ \Big]=\int \frac{\dfrac{2\, dt}{1+t^2}}{\dfrac{1-t^2}{1+t^2}}=\\\\\\=\int \frac{2\, dt}{1-t^2}=2\cdot \frac{1}{2}ln\left|\, \frac{1+t}{1-t}\, \right|+C=ln\left|\, \frac{1+tg\dfrac{u}{2}}{1-tg\dfrac{u}{2}}\, \right|+C=\\\\\\=ln\left|\, \dfrac{tg\dfrac{\pi}{4}+tg\dfrac{u}{2}}{1-tg\dfrac{\pi }{4}\cdot tg\dfrac{u}{2}}\, \right|+C=ln\Big|\, tg\Big(\dfrac{u}{2}+\dfrac{\pi}{4}\Big)\, \Big|+C$