Найдите,пожалуйста,неопределенный интеграл!

Question 1

Question 2

Решите задачу:

$\int\limits {\frac{dx}{sinx\sqrt{1+cosx}} \, dx =[\, cos^2\frac{x}{2}=\frac{1+cosx}{2},\sqrt{1+cosx}=\sqrt{2cos^2\frac{x}{2}}=\sqrt2cos\frac{x}{2}\, ]=$

$=\int \frac{dx}{sinx\cdot \sqrt2\cdot cos\frac{x}{2}} =\int \frac{dx}{2sin\frac{x}{2}\cdot cos\frac{x}{2}\cdot \sqrt2cos\frac{x}{2}}=\frac{1}{2\sqrt2}\int \frac{(sin^2\frac{x}{2}+cos^2\frac{x}{2})dx}{sin\frac{x}{2}\cdot cos^2\frac{x}{2}} =\\\\=\frac{1}{2\sqrt2}\int (\frac{sin\frac{x}{2}}{cos^2\frac{x}{2}}+\frac{1}{sin\frac{x}{2}})dx=\frac{1}{2\sqrt2}\int \frac{sin\frac{x}{2}dx}{cos^2\frac{x}{2}}+\frac{1}{2\sqrt2}\int \frac{sin\frac{x}{2}dx}{sin^2\frac{x}{2}} =$

$=\frac{1}{2\sqrt2}\int \frac{-2dt}{t^2}\; \; [\, t=cos\frac{x}{2},dt=-\frac{1}{2}sin\frac{x}{2}dx\, ]+\frac{1}{2\sqrt2}\int \frac{sin\frac{x}{2}dx}{1-cos^2\frac{x}{2}}=\\\\=-\frac{1}{\sqrt2}\cdot \frac{t^{-1}}{-1} +\frac{1}{2\sqrt2}\int \frac{-2dt}{1-t^2} =\frac{1}{\sqrt2\, t}-\frac{1}{\sqrt2}\int \frac{dt}{-(t^2-1)}=\\\\=\frac{1}{\sqrt2\, t}+\frac{1}{\sqrt2}\cdot \frac{1}{2}\cdot ln\left |\frac{t-1}{t+1} \right |+C=$

$=\frac{1}{\sqrt2cos\frac{x}{2}}+\frac{1}{2\sqrt2}\cdot ln\left |\frac{cos\frac{x}{2}-1}{cos\frac{x}{2}+1}\right |+C$