int{ (cos(x)+sin(x)) / (5cos^2(x)-2sin(2x)+2sin^2(x)) dx}

#1

Νά αποδειχθεί ότι:

$\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\frac{3}{5}\,\arctan\left({\sin{x}-2\,\cos{x}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\right|$

ΔΙΟΡΘΩΣΗ [29-4-09 11:10 μ.μ.] : Τό αρχικό αποτέλεσμα πού έδωσα, άν καί προέρχεται από Πανεπιστημιακές σημειώσεις, ήταν λανθασμένο - συμβαίνει παντού - καί τό διόρθωσα.

#2

καί η επίλυση πού βρήκα...

$I=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-4\,\sin{x}\,\cos{x}+2\,\sin^2{x}}\,dx}.$

Θέτωντας $t=\tan{x}$ , προκύπτουν $\sin{x}=\displaystyle\frac{t}{\sqrt{1+t^2}}$ , $\cos{x}=\displaystyle\frac{1}{\sqrt{1+t^2}}$ καί $dx=\displaystyle\frac{1}{1+t^2}\,dt$ .

$I=\displaystyle\int{\frac{\frac{1}{\sqrt{1+t^2}}+\frac{t}{\sqrt{1+t^2}}}{5\,\frac{1}{1+t^2}-4\,\frac{t}{1+t^2}+2\,\frac{t^2}{1+t^2}}\,\frac{1}{1+t^2}\,dt}=\int{\frac{1+t}{\left({5-4t+2t^2}\right)\sqrt{1+t^2}}\,dt}=$

$\displaystyle\int{\frac{\frac{3}{5}({2t+1})-\frac{\sqrt{6}}{30}\,\sqrt{6}\,({t-2})}{\left({2t^2-4t+5}\right)\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{2t+1}{\left({({t-2})^2+t^2+1}\right)\sqrt{1+t^2}}\,dt}-\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t-2})}{\left({\left({\sqrt{6}\,\sqrt{1+t^2}}\right)^2-({2t+1})^2}\right)\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{2t+1}{\left({\left({\frac{t-2}{\sqrt{1+t^2}}}\right)^2+1}\right)\left({1+t^2}\right)^{\frac{3}{2}}}\,dt}-\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t-2})}{\left({\left({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)^2-1}\right)({2t+1})^2\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{1}{\left({\frac{t-2}{\sqrt{1+t^2}}}\right)^2+1}\,d\!\left({\tfrac{t-2}{\sqrt{1+t^2}}}\right)}-\frac{\sqrt{6}}{30}\int{\frac{1}{\left({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)^2-1}\,d\!\left({\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)}=$

$\displaystyle\frac{3}{5}\int{\frac{1}{\left({\frac{t-2}{\sqrt{1+t^2}}}\right)^2+1}\,d\!\left({\tfrac{t-2}{\sqrt{1+t^2}}}\right)}-\frac{\sqrt{6}}{30}\int{\frac{1}{\left({\left({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)-1}\right)\left({\left({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)+1}\right)}\,d\!\left({\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\right)}=$

$\displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t-2}{\sqrt{1+t^2}}}\right)-\frac{\sqrt{6}}{60}\left({\ln\left|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}-1}\right|-\ln\left|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}+1}\right|}\right)+c=$

$\displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t-2}{\sqrt{1+t^2}}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}+2t+1}{\sqrt{6}\,\sqrt{1+t^2}-2t-1}}\right|+c=$

$\displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t}{\sqrt{1+t^2}}-2\,\tfrac{1}{\sqrt{1+t^2}}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\frac{\sqrt{6}\,\sqrt{1+t^2}+2t+1}{\sqrt{1+t^2}}}{\frac{\sqrt{6}\,\sqrt{1+t^2}-2t-1}{\sqrt{1+t^2}}}}\right|+c=$

$\displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t}{\sqrt{1+t^2}}-2\,\tfrac{1}{\sqrt{1+t^2}}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\sqrt{6}+2\,\frac{t}{\sqrt{1+t^2}}+\frac{1}{\sqrt{1+t^2}}}{\sqrt{6}-2\,\frac{t}{\sqrt{1+t^2}}-\frac{1}{\sqrt{1+t^2}}}}\right|+c\,\stackrel{t=\tan{x}}{=}$

$\displaystyle\frac{3}{5}\,\arctan\left({\sin{x}-2\,\cos{x}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\right|+c.\quad\square$

int{ (cos(x)+sin(x)) / (5cos^2(x)-2sin(2x)+2sin^2(x)) dx}

int{ (cos(x)+sin(x)) / (5cos^2(x)-2sin(2x)+2sin^2(x)) dx}

Re: int{ (cos(x)+sin(x)) / (5cos^2(x)-2sin(2x)+2sin^2(x)) dx}

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