1/Cos x'in İntegrali

1/Cos x'in İntegrali Nedir ?

1/Cos x'in integrali ln |sec x+tan x|+c'dir.

$\int \frac{dx}{cosx}=ln|secx+tanx|+c$

$\int \frac{dx}{cosx}=-ln|secx-tanx|+c$

$\int \frac{dx}{cosx}=\frac{1}{2}.ln|\frac{1+sinx}{1-sinx}|+c$

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

1/Cos x'in İntegralini Bulma

$\boxed{\frac{1}{cosx}=secx}$

$\int \frac{1}{cosx}dx=\int secxdx$

$\int \frac{dx}{cosx}=\int secxdx$

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

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

$secx+tanx=u$

$d(secx+tanx)=du$

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

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

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

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

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

$\int \frac{dx}{cosx}=\int \frac{du}{u}$

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

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

$\int \frac{dx}{cosx}=ln|secx+tanx|+c$

$secx+tanx=(\frac{1}{secx+tanx}{)}^{-1}$

$\int \frac{dx}{cosx}=ln|(\frac{1}{secx+tanx}{)}^{-1}|+c$

$\int \frac{dx}{cosx}=-ln|\frac{1}{secx+tanx}|+c$

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

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

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

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

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

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

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

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

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

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

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

$\int \frac{dx}{cosx}=-ln|secx-tanx|+c$

$-ln|secx-tanx|=ln|(secx-tanx{)}^{-1}|$

$\int \frac{dx}{cosx}=ln|(secx-tanx{)}^{-1}|+c$

$\int \frac{dx}{cosx}=ln|\frac{1}{secx-tanx}|+c$

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

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

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

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

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

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

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

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

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

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

$\int \frac{dx}{cosx}=\frac{1}{2}.ln|\frac{1+sinx}{1-sinx}|+c$

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

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

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

$\boxed{sin2x=2.sinx.cosx}$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+2.sin\frac{x}{2}.cos\frac{x}{2}}{1-2.sin\frac{x}{2}.cos\frac{x}{2}}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+2.\frac{t}{\sqrt{1+t^2}}.\frac{1}{\sqrt{1+t^2}}}{1-2.\frac{t}{\sqrt{1+t^2}}.\frac{1}{\sqrt{1+t^2}}}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+\frac{2.t.1}{\sqrt{1+t^2}.\sqrt{1+t^2}}}{1-\frac{2.t.1}{\sqrt{1+t^2}.\sqrt{1+t^2}}}}|+c$

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

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

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+\frac{2.t}{1+t^2}}{1-\frac{2.t}{1+t^2}}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{\frac{1+t^2+2.t}{1+t^2}}{\frac{1+t^2-2.t}{1+t^2}}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{\frac{1+t^2+2.t}{\cancel{1+t^2}}}{\frac{1+t^2-2.t}{\cancel{1+t^2}}}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+t^2+2.t}{1+t^2-2.t}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1+2.t+t^2}{1-2.t+t^2}}|+c$

$\int \frac{dx}{cosx}=ln|\sqrt{\frac{1^2+2.t+t^2}{1^2-2.t+t^2}}|+c$

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

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

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

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

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

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

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

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

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