Question

Решить неопределенный интеграл

Answer 1

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

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

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