1/Cos x'in İntegrali

1/Cos x'in İntegrali Nedir ?

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

1/Cos x'in İntegralini Bulma

1. Yol

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

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

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

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

$secx+tanx=u$

$d(secx+tanx)=du$

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

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

$\boxed{(secx{)}^{\prime }=secx.tanx}$ $\boxed{(tanx{)}^{\prime }=1+ta{n}^{2}x=\frac{1}{co{s}^{2}x}=se{c}^{2}x}$

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

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

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

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

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

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

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

$\int \frac{dx}{cosx}=ln|\frac{se{c}^{2}x-ta{n}^{2}x}{secx-tanx}|+c$

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

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

$\int \frac{dx}{cosx}=ln|\frac{\frac{1}{co{s}^{2}x}-\frac{si{n}^{2}x}{co{s}^{2}x}}{secx-tanx}|+c$

$\int \frac{dx}{cosx}=ln|\frac{\frac{1-si{n}^{2}x}{co{s}^{2}x}}{secx-tanx}|+c$

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

$\int \frac{dx}{cosx}=ln|\frac{\frac{\cancel{co{s}^{2}x}}{\cancel{co{s}^{2}x}}}{secx-tanx}|+c$

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

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

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

2. Yol

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

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

$\int \frac{dx}{cosx}=\int \frac{cosx}{co{s}^{2}x}dx$

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

$\int \frac{dx}{cosx}=\int \frac{cosx}{1-si{n}^{2}x}dx$

$sinx=u$

$d(sinx)=du$

$(sinx{)}^{\prime }dx=du$

$\boxed{(sinx{)}^{\prime }=cosx}$

$cosxdx=du$

$\int \frac{dx}{cosx}=\int \frac{du}{1-u^2}$

$\int \frac{dx}{cosx}=\int \frac{du}{(1-u).(1+u)}$

$\frac{1}{(1-u).(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

$\frac{1}{(1-u).(1+u)}=\frac{A.(1+u)+B.(1-u)}{(1-u).(1+u)}$

$\frac{1}{\cancel{(1-u).(1+u)}}=\frac{A+Au+B-Bu}{\cancel{(1-u).(1+u)}}$

$0.u+1=(A-B).u+(A+B)$

$A-B=0$

$A+B=1$

$A-\cancel{B}+A+\cancel{B}=0+1$

$2A=1$

$A=\frac{1}{2}$

$\frac{1}{2}+B=1$

$B=1-\frac{1}{2}=\frac{1}{2}$

$\int \frac{dx}{cosx}=\int \frac{1/2}{1-u}du+\int \frac{1/2}{1+u}du$

$\int \frac{dx}{cosx}=\frac{1}{2}\int \frac{du}{1-u}+\frac{1}{2}\int \frac{du}{1+u}$

$\int \frac{dx}{cosx}=\frac{1}{2}(\int \frac{du}{1-u}+\int \frac{du}{1+u})$

$\boxed{\int \frac{dx}{a\pm x}=\pm ln|a\pm x|}$

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

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

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

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

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

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