ολοκλήρωμα tan^(1/2)(x)

#1

$\displaystyle\int{\sqrt{\tan{x}}\,dx}$

[ έχω κάνει μιά αρκετά μεγάλη επίλυση ]

mathxl · #2

Μπορούμε να δούμε εδώ μία προσέγγιση http://www.mathlinks.ro/viewtopic.php?t=20165

#3

mathxl έγραψε:Μπορούμε να δούμε εδώ μία προσέγγιση ....mathlinks.ro/viewtopic.php?t=20165

Ειλικρινά τούς έχω "ξεχάσει" τούς mathlinkers! Πάντως έχω κάνει τήν ίδια προσέγγιση...

#4

Θέτωντας $t=\tan{x}$ , προκύπτει $dx=\dfrac{1}{1+t^2}\,dt$ .

$\displaystyle\int{\sqrt{\tan{x}}\,dx}=\int{\frac{\sqrt{t}}{1+t^2}\,dt}\stackrel{u^2\,=\,t}{=}\int{\frac{2u^2}{1+u^4}\,du}=\int{\frac{2u^2}{\left({1+u^2}\right)^2-\left({u\,\sqrt{2}}\right)^2}\,du}=\int{\frac{2u^2}{\left({1+u\,\sqrt{2}+u^2}\right)\left({1-u\,\sqrt{2}+u^2}\right)}\,du}=$

$\displaystyle\int{\frac{-\frac{\sqrt{2}}{2}\,u}{1+u\,\sqrt{2}+u^2}\,du}+\int{\frac{\frac{\sqrt{2}}{2}\,u}{1-u\,\sqrt{2}+u^2}\,du}=$

$\displaystyle-\frac{\sqrt{2}}{4}\int{\frac{2u+\sqrt{2}-\sqrt{2}}{1+u\,\sqrt{2}+u^2}\,du}+\frac{\sqrt{2}}{4}\int{\frac{2u-\sqrt{2}+\sqrt{2}}{1-u\,\sqrt{2}+u^2}\,du}=$

$\displaystyle-\frac{\sqrt{2}}{4}\int{\frac{\left({1+u\,\sqrt{2}+u^2}\right)^{\prime}}{1+u\,\sqrt{2}+u^2}\,du}+\frac{1}{2}\int{\frac{1}{1+u\,\sqrt{2}+u^2}\,du}+\frac{\sqrt{2}}{4}\int{\frac{\left({1-u\,\sqrt{2}+u^2}\right)^{\prime}}{1-u\,\sqrt{2}+u^2}\,du}+\frac{1}{2}\int{\frac{1}{1-u\,\sqrt{2}+u^2}\,du}=$

$\displaystyle-\frac{\sqrt{2}}{4}\log\left|{1+u\,\sqrt{2}+u^2}\right|+\int{\frac{1}{1+\left({1+u\,\sqrt{2}}\right)^2}\,du}+\frac{\sqrt{2}}{4}\log\left|{1-u\,\sqrt{2}+u^2}\right|+\int{\frac{1}{1+\left({1-u\,\sqrt{2}}\right)^2}\,du}+c_1=$

$\displaystyle\frac{\sqrt{2}}{4}\log\left|{\tfrac{1-u\,\sqrt{2}+u^2}{1+u\,\sqrt{2}+u^2}}\right|+\frac{\sqrt{2}}{2}\arctan\left({1+u\,\sqrt{2}}\right)+\frac{\sqrt{2}}{2}\arctan\left({1-u\,\sqrt{2}}\right)+c\stackrel{u\,=\,\sqrt{\tan{x}}}{=}$

$\displaystyle\frac{\sqrt{2}}{4}\log\left|{\tfrac{1-\sqrt{2\,\tan{x}}+\tan{x}}{1+\sqrt{2\,\tan{x}}+\tan{x}}}\right|+\frac{\sqrt{2}}{2}\arctan\left({1+\sqrt{2\,\tan{x}}}\right)+\frac{\sqrt{2}}{2}\arctan\left({1-\sqrt{2\,\tan{x}}}\right)+c=\frac{\sqrt{2}}{4}\log\left|{\tfrac{1-\sqrt{2\,\tan{x}}+\tan{x}}{1+\sqrt{2\,\tan{x}}+\tan{x}}}\right|+\frac{\sqrt{2}}{2}\arctan\tfrac{1}{\tan{x}}+c.\quad\square$

ολοκλήρωμα tan^(1/2)(x)

ολοκλήρωμα tan^(1/2)(x)

Re: ολοκλήρωμα tan^(1/2)(x)

Re: ολοκλήρωμα tan^(1/2)(x)

Re: ολοκλήρωμα tan^(1/2)(x)

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