$\int \frac{dx}{tgx\cdot cos2x}=\int \frac{dx}{\frac{sinx}{cosx}\cdot (cos^2x-sin^2x)}=\int \frac{cosx\, dx}{sinx\cdot cos^2x-sin^3x}=\Big [\frac{:cos^3x}{:cos^3x}\Big ]=\\\\=\int \frac{\frac{dx}{cos^2x}}{tgx-tg^3x}=\int \frac{d(tgx)}{-tgx\cdot (tg^2x-1)}=\Big [\; t=tgx\; ,\; dt=\frac{dx}{cos^2x}\; \Big ]=\\\\=-\int \frac{dt}{t\cdot (t-1)(t+1)}=I\\\\\\\frac{1}{t(t-1)(t+1)}=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+1}\\\\1=A(t-1)(t+1)+Bt(t+1)+Ct(t-1)\\\\t=0:\; A=\frac{1}{-1}=-1\\\\t=1:\; \; B=\frac{1}{1\cdot 2}=\frac{1}{2}$

$t=-1:\; \; C=\frac{1}{-1\cdot (-2)}=\frac{1}{2}\\\\\\I=-\int \frac{-1}{t}\, dt-\int \frac{1}{2(t-1)}\, dt-\int \frac{1}{2(t+1)}\, dt=\\\\=ln\, |t|-\frac{1}{2}ln|t-1|-\frac{1}{2}ln|t+1|+C=\\\\=ln\, |tgx|-\frac{1}{2}\cdot ln\Big |tgx-1)(tgx+1)\Big |+C=ln\, |tgx|-\frac{1}{2}\cdot ln\Big |tg^2x-1\Big |+C$

$2)\; \; \int \frac{\sqrt[4]{x}+1}{(\sqrt{x}+4)\cdot \sqrt[4]{x^3}}\, dx=\Big [\; x=t^4\; ,\; dx=4t^3\, dt\; ,\; \sqrt[4]{x}=t\; ,\\\\\sqrt[4]{x^3}=t^3\; ,\; \sqrt{x}=t^2\; \Big ]=\int \frac{(t+1)\cdot 4t^3\, dt}{(t^2+4)\cdot t^3}=4\int \frac{(t+1)dt}{t^2+4}=\\\\=2\int \frac{2t\, dt}{t^2+4}+4\int \frac{dt}{t^2+4}=2\int \frac{d(t^2+4)}{t^2+4}+4\cdot \frac{1}{2}\, arctg\frac{t}{2}=\\\\=2\, ln(t^2+4)+2\, arctg\frac{t}{2}+C=2\cdot ln(\sqrt{x}+4)+2\, arctg\frac{\sqrt[4]{x}}{2}+C$

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3 года назад

A1=1 An+1= An+3 И A1=2 An+1=3An Найти А5 рекуррентно

Юрий78

$1)\; \; a_1=1\; ,\; \; a_{n+1}=a_{n}+3\\\\a_2=a_1+3=1+3=4\; ,\\\\a_3=a_2+3=4+3=7\\\\a_4=a_3+3=7+3=10\\\\a_5=a_4+3=10+3=13\\\\ili\\\\a_5=a_4+3=(a_3+3)+3=a_3+2\cdot 3=(a_2+3)+2\cdot 3=\\\\=a_2+3\cdot 3=(a_1+3)+3\cdot 3=a_1+4\cdot 3=1+4\cdot 3=13\\\\\\2)\; \; a_1=2\; ,\; \; a_{n+1}=3\, a_{n}\\\\a_2=3\, a_1=3\cdot 2=6\\\\a_3=3\, a_2=3\cdot 6=18\\\\a_4=3\, a_3=3\cdot 18=54\\\\a_5=3\, a_4=3\cdot 54=162$