Question

Вычислить интеграл
Помогите пожалуйста !!!!!!!

Answer 1

$\LARGE \int_{0}^{\pi\over2}{\sin{x}\over 1+\cos{x}}\mathrm{dx}=-\int_{0}^{\pi\over2}{\mathrm{d(1+\cos{x})}\over1+\cos{x}}=-\ln{(|1+\cos{x}|)}|_{0}^{\pi\over2}=-(\ln{1}-\ln{2})=\ln{2}$

Answer 2

$\displaystyle \int\limits^\frac{\pi}{2}_0\frac{sinxdx}{1+cosx}=-\int\limits^\frac{\pi}{2}_0\frac{d(1+cosx)}{1+cosx}=-ln|1+cosx||^\frac{\pi}{2}_0=-ln1+ln2\approx\\\approx0,693\\\\\\\\\displaystyle \int\limits^\frac{\pi}{2}_0\frac{sinxdx}{1+cosx}=\int\limits^2_1\frac{dt}{t}=ln|t||^2_1=ln2-ln1=ln2\approx0,693\\t=1+cosx;dt=-sinxdx;dx=-\frac{dt}{sinx}\\t_1=1;t_2=2$