Интегралы помогите под 27 номером

Question 1

Question 2

Решите задачу:

$1)\; \; \int 2sin^2 \frac{x}{2}\, dx=2\int\frac{1-cosx}{2}\, dx=\int (1-cosx)dx=x-sinx+C\\\\2)\; \; \int \frac{dx}{sin^3x}=\int \frac{dx}{(2\, sin\frac{x}{2}\cdot cos\frac{x}{x})^3}=\frac{1}{8}\int\frac{dx}{\frac{sin^3\frac{x}{2}\cdot cos^3\frac{x}{2}}{cos^6\frac{x}{2}}\cdot cos^6\frac{x}{2}}=\\\\=\frac{1}{8}\int \frac{1}{tg^3\frac{x}{2}}\cdot \frac{1}{cos^2\frac{x}{2}}\cdot \frac{1}{cos^2\frac{x}{2}}\cdot \frac{dx}{cos^2\frac{x}{2}}=[\, 1+tg^2 \alpha =\frac{1}{cos^2 \alpha }\, ]=\\\\=[\, t=tg\frac{x}{2}\; ,\; d(tg\frac{x}{2})=\frac{dx}{2cos^2\frac{x}{2}}\, ]=\frac{2}{8}\int \, \frac{1}{t^{3}}\cdot (1+t^2)^2\cdot dt=$

$=\frac{1}{4}\int \frac{1+2t^2+t^4}{t^3}\, dt=\frac{1}{4}\int (t^{-3}+\frac{2}{t} +t)dt=\\\\=\frac{1}{4}\cdot (\frac{t^{-2}}{-2}+2\, ln|t|+\frac{t^2}{2})+C=\frac{1}{4}\cdot (-\frac{1}{2t^2}+2\, ln|t|+\frac{t^2}{2})+C=\\\\=-\frac{1}{8tg^2\frac{x}{2}}+\frac{1}{2}\cdot ln|tg \frac{x}{2}|+\frac{1}{8}\cdot tg^2\frac{x}{2}+C$

$3)\; \; \int \frac{e^{z}\sqrt{arctge^{z}}}{1+e^{2z}}\, dz=[\, t=e^{z},\; z=lnt\; ,\; dz=\frac{dt}{t}\, ]=\\\\=\int \frac{t\cdot \sqrt{atctgt}}{1+t^2}\cdot \frac{dt}{t}=\int \frac{\sqrt{arctgt}}{1+t^2}\cdot dt=[\, d(arctgt)=\frac{dt}{1+t^2}\, ]=\\\\=\int \sqrt{arctgt}\cdot d(arctgt)=\frac{(arctgt)^{3/2}}{3/2}+C=\frac{2}{3}\cdot \sqrt{arctg^3t}+C=\\\\=\frac{2}{3}\cdot \sqrt{arctg^3e^{z}}+C$