Tan x'in İntegrali

Tan x'in İntegrali Nedir ?

Tan x'in integrali -ln |cos x|+c'dir.

$\int tanxdx=-ln|cosx|+c$

$\int tanxdx=ln|secx|+c$

Tan x'in İntegralini Bulma

1. Yol

$\int tanxdx=?$

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

$\int tanxdx=\int \frac{sinx}{cosx}dx$

$\int tanxdx=\int \frac{sinxdx}{cosx}$

$cosx=u$

$d(cosx)=d(u)$

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

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

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

$-sinxdx=du$

$sinxdx=-du$

$\int tanxdx=\int \frac{-du}{u}$

$\int tanxdx=-\int \frac{du}{u}$

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

$\int tanxdx=-ln|u|+c$

$\int tanxdx=-ln|cosx|+c$

2. Yol

$\int tanxdx=\int \frac{secx.tanxdx}{secx}$

$secx=u$

$d(secx)=d(u)$

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

$(secx{)}^{\prime }=secx.tanx$

$secx.tanxdx=du$

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

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

$\int tanxdx=ln|secx|+c$

3. Yol

$\int tanxdx=?$

$tanx=u$

$d(tanx)=d(u)$

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

$(tanx{)}^{\prime }=1+ta{n}^{2}x$

$(1+ta{n}^{2}x)dx=du$

$(1+u^2)dx=du$

$dx=\frac{du}{1+u^2}$

$\int tanxdx=\int u\frac{du}{1+u^2}$

$\int tanxdx=\int \frac{udu}{1+u^2}$

$1+u^2=v$

$d(1+u^2)=d(v)$

$(1+u^2{)}^{\prime }du=dv$

$\frac{\cancel{2}udu}{\cancel{2}}=\frac{dv}{2}$

$udu=\frac{dv}{2}$

$\int tanxdx=\int \frac{\frac{dv}{2}}{v}$

$\int tanxdx=\int \frac{\frac{1}{2}dv}{v}$

$\int tanxdx=\frac{1}{2}\int \frac{dv}{v}$

$\int tanxdx=\frac{1}{2}ln|v|+c$

$\int tanxdx=ln|v{|}^{\frac{1}{2}}+c$

$\int tanxdx=ln|v^{\frac{1}{2}}|+c$

$\int tanxdx=ln|\sqrt{v}|+c$

$\int tanxdx=ln|\sqrt{1+u^2}|+c$

$tanx=u$

$secx=\sqrt{1+u^2}$

$\int tanxdx=ln|secx|+c$