Question

помогите решить определённый интеграл

Answer 1

$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}} {\frac{x\, dx}{sin^2x}} \, dx =[\, u=x,\; du=dx,\; dv=\frac{dx}{sin^2x},\; v=-ctgx\, ]=\\\\=-x\cdot ctgx|_{\frac{\pi}{4}}^{\frac{\pi}{2}}+ \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}} {ctgx} \, dx =-\frac{\pi}{2}\cdot 0+\frac{\pi}{4}\cdot ctg\frac{\pi}{4}+ \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}} {\frac{cosx\, dx}{sinx}} =\\\\=[z=sinx,\; dz=cosx\, dx,\int \frac{dz}{z}=ln|z|+C\, ]=\frac{\pi}{4}+ln|sinx|\, |_{\frac{\pi}{4}}^{\frac{\pi}{2}}=$

$=\frac{\pi}{4}+ln|sin\frac{\pi}{2}|-ln|sin\frac{\pi}{4}|=\frac{\pi}{4}+ln1-ln\frac{\sqrt2}{2}=\frac{\pi}{4}-ln2^{-\frac{1}{2}}=\\\\=\frac{\pi}{4}+\frac{1}{2}ln2=\frac{1}{4}\cdot (\pi +2ln2)=\frac{1}{4}\cdot (\pi +ln4)$