Question

Решение определенного интеграла. Срочно

Answer 1

$1)\; \; \int\limits^{\pi /3}_{\pi /4}\, \frac{x\, dx}{cos^2x}}=[\, u=x,\; du=dx,\; dv=\frac{dx}{cos^2x},\; v=tgx\, ]=\\\\=x\, tgx\Big |_{\pi /4}^{\pi /3}-\int\limits^{\pi /3}_{\pi /4}\, tgx\, dx=\frac{\pi}{3}\cdot tg\frac{\pi}{3}-\frac{\pi}{4}\cdot tg\frac{\pi}{4}- \int\limits^{\pi /3}_{\pi /4} \frac{sinx}{cosx}\, dx=\\\\=\frac{\pi}{3}\cdot \sqrt3-\frac{\pi}{4}+\int\limits^{\pi /3}_{\pi /4}\, \frac{d(cosx)}{cosx}=[\, \int \frac{dt}{t}=ln|t|+C\; ,\; d(cosx)=-sinx\, dx\, ]=\\\\=\frac{\pi \sqrt3}{3}-\frac{\pi }{4}+ln|cosx|\Big |_{\pi /4}^{\pi /3}=$

$=\frac{\pi \sqrt3}{3}-\frac{\pi}{4}+(ln\frac{1}{2}-ln\frac{\sqrt2}{2})=\frac{\pi \sqrt3}{3}-\frac{\pi }{4}+ln\sqrt2$

$2)\; \; \int\limits^4_0\, \frac{4x+3}{\sqrt{2x+1}+5}\, dx=[\, t^2=2x+1,\; x=\frac{t^2-1}{2},\; dx=t\, dt\, ]=\\\\=\int\limits^3_1\, \frac{2t^2+1}{t+5}\, dt =\int\limits^3_1\, (2t-10+\frac{51}{t+5})dt=(2\cdot \frac{t^2}{2}-10t+51\cdot ln|t+5|)\Big |_1^3=$

$=9-30+51ln8-(1-10+ln6)=-12+51\, ln\frac{4}{3}$