Sec x'in İntegrali

Sec x'in İntegrali Nedir ?

Sec x'in integrali ln |sec x+tan x|+c'dir.

$\int secxdx=ln|secx+tanx|+c$

$\int secxdx=-ln|secx-tanx|+c$

$\int secxdx=\frac{1}{2}ln|\frac{1+sinx}{1-sinx}|+c$

$\int secxdx=ln|tan(\frac{x}{2}+\frac{\pi }{4})|+c$

Sec x'in İntegralini Bulma

$\int secxdx=\int \frac{(secx+tanx).secxdx}{secx+tanx}$

$\int secxdx=\int \frac{secx.secxdx+secx.tanxdx}{secx+tanx}$

$\int secxdx=\int \frac{se{c}^{2}xdx+secx.tanxdx}{secx+tanx}$

$\int secxdx=\int \frac{(se{c}^{2}x+secx.tanx)dx}{secx+tanx}$

$secx+tanx=u$

$d(secx+tanx)=d(u)$

$d[f(x)]=f^{\prime }(x)dx$

$(secx+tanx{)}^{\prime }dx=du$

$(secx{)}^{\prime }dx+(tanx{)}^{\prime }dx=du$

$(secx{)}^{\prime }=secx.tanx$ $(tanx{)}^{\prime }=se{c}^{2}x$

$secx.tanxdx+se{c}^{2}xdx=du$

$(secx.tanx+se{c}^{2}x)dx=du$

$\int secxdx=\int \frac{du}{u}$

$\int \frac{dx}{x}=ln|x|+c$

$\int secxdx=ln|u|+c$

$\int secxdx=ln|secx+tanx|+c$

$secx=\frac{1}{cosx}$ $tanx=\frac{sinx}{cosx}$

$\int secxdx=ln|\frac{1}{cosx}+\frac{sinx}{cosx}|+c$

$\int secxdx=ln|\frac{1+sinx}{cosx}|+c$

$\int secxdx=ln|\frac{(1-sinx).(1+sinx)}{(1-sinx).cosx}|+c$

$\int secxdx=ln|\frac{1-si{n}^{2}x}{(1-sinx).cosx}|+c$

$co{s}^{2}x=1-si{n}^{2}x$

$\int secxdx=ln|\frac{\cancel{co{s}^{2}x}cosx}{(1-sinx).\cancel{cosx}}|+c$

$\int secxdx=ln|\frac{cosx}{1-sinx}|+c$

$\int secxdx=ln|(\frac{1-sinx}{cosx}{)}^{-1}|+c$

$\int secxdx=ln|(\frac{1-sinx}{cosx}){|}^{-1}+c$

$\int secxdx=-ln|(\frac{1-sinx}{cosx})|+c$

$\int secxdx=-ln|\frac{1}{cosx}-\frac{sinx}{cosx}|+c$

$\int secxdx=-ln|secx-tanx|+c$

$\int secxdx=ln|secx+tanx|+c$

$\int secxdx=ln|\frac{1}{cosx}+\frac{sinx}{cosx}|+c$

$\int secxdx=ln|\frac{1+sinx}{cosx}|+c$

$cosx=\sqrt{1-si{n}^{2}x}$

$\int secxdx=ln|\frac{\sqrt{(1+sinx{)}^2}}{\sqrt{1-si{n}^{2}x}}|+c$

$\int secxdx=ln|\frac{\sqrt{(1+sinx).(1+sinx)}}{\sqrt{(1-sinx).(1+sinx)}}|+c$

$\int secxdx=ln|\sqrt{\frac{(1+sinx).\cancel{(1+sinx)}}{(1-sinx).\cancel{(1+sinx)}}}|+c$

$\int secxdx=ln|\sqrt{\frac{1+sinx}{1-sinx}}|+c$

$\int secxdx=ln|(\frac{1+sinx}{1-sinx}{)}^{\frac{1}{2}}|+c$

$\int secxdx=ln|\frac{1+sinx}{1-sinx}{|}^{\frac{1}{2}}+c$

$\int secxdx=\frac{1}{2}ln|\frac{1+sinx}{1-sinx}|+c$

$\int secxdx=ln|secx+tanx|+c$

$\int secxdx=ln|\frac{1}{cosx}+\frac{sinx}{cosx}|+c$

$\int secxdx=ln|\frac{1+sinx}{cosx}|+c$

$sin2x=2.sinx.cosx$ $cos2x=co{s}^{2}x-si{n}^{2}x$

$\int secxdx=ln|\frac{1+2.sin\frac{x}{2}.cos\frac{x}{2}}{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}|+c$

$tan\frac{x}{2}=t$

$sin\frac{x}{2}=\frac{t}{\sqrt{1+t^2}}$

$cos\frac{x}{2}=\frac{1}{\sqrt{1+t^2}}$

$\int secxdx=ln|\frac{1+2.\frac{t}{\sqrt{1+t^2}}.\frac{1}{\sqrt{1+t^2}}}{(\frac{1}{\sqrt{1+t^2}}{)}^2-(\frac{t}{\sqrt{1+t^2}}{)}^2}|+c$

$\int secxdx=ln|\frac{1+\frac{2.t.1}{\sqrt{1+t^2}.\sqrt{1+t^2}}}{\frac{1}{(\sqrt{1+t^2}{)}^2}-\frac{t^2}{(\sqrt{1+t^2}{)}^2}}|+c$

$\int secxdx=ln|\frac{1+\frac{2t}{(\sqrt{1+t^2}{)}^2}}{\frac{1}{(\sqrt{1+t^2}{)}^2}-\frac{t^2}{(\sqrt{1+t^2}{)}^2}}|+c$

$\int secxdx=ln|\frac{1+\frac{2t}{1+t^2}}{\frac{1}{1+t^2}-\frac{t^2}{1+t^2}}|+c$

$\int secxdx=ln|\frac{\frac{1+t^2+2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}|+c$

$\int secxdx=ln|\frac{\frac{1+2t+t^2}{\cancel{1+t^2}}}{\frac{1-t^2}{\cancel{1+t^2}}}|+c$

$\int secxdx=ln|\frac{1+2t+t^2}{1-t^2}|+c$

$\int secxdx=ln|\frac{(1+t{)}^2}{(1-t).(1+t)}|+c$

$\int secxdx=ln|\frac{(1+t).\cancel{(1+t)}}{(1-t).\cancel{(1+t)}}|+c$

$\int secxdx=ln|\frac{1+t}{1-t}|+c$

$\int secxdx=ln|\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}|+c$

$\int secxdx=ln|\frac{1+tan\frac{x}{2}}{1-1.tan\frac{x}{2}}|+c$

$tan\frac{\pi }{4}=1$

$\int secxdx=ln|\frac{tan\frac{\pi }{4}+tan\frac{x}{2}}{1-tan\frac{\pi }{4}.tan\frac{x}{2}}|+c$

$tan(p+q)=\frac{tanp+tanq}{1-tanp.tanq}$

$\int secxdx=ln|tan(\frac{\pi }{4}+\frac{x}{2})|+c$