Question

qual a integral da funcao cossec x dx

Answer 1

$\int\ {cscx} \, dx$

$\int\ {\frac{1}{sinx}} \, dx$

$\int\ {\frac{sinx}{sin^2x}} \, dx$

$\int\ {\frac{sinx}{1-cos^2x}} \, dx$

$u=cosx$

$du=-sinx dx$

$dx=-\frac{du}{sinx}$

$- \int\ {\frac{sinx}{(1-u^2)sinx}} \, du$

$- \int\ {\frac{1}{1-u^2}} \, du$

$- \int\ {\frac{1}{(1-u)(1+u)}} \, du$

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

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

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

$\left \{ {{A+B=1} \atop {A-B=0}} \right.$

$A=\frac12, B=\frac12$

$\frac12\frac{1}{1-u}+\frac12\frac{1}{1+u}=\frac{1}{(1-u)(1+u)}$

$- \int\ {\frac12\frac{1}{1-u}+\frac12\frac{1}{1+u}} \, du$

$-( \int\ {\frac12\frac{1}{1-u}} \, du+ \int\ {\frac12\frac{1}{1+u}} \, du )$

$-(\frac12 \int\ {\frac{1}{1-u}} \, du+ \frac12\int\ {\frac{1}{1+u}} \, du )$

$-(\frac12 (-ln|1-u|)+ \frac12(ln|1+u| ))$

$-\frac12( (-ln|1-u|)+(ln|1+u| ))$

$-\frac12 ln|\frac{1+u}{1-u}|$

$u=cosx$

$-\frac12 ln|\frac{1+cosx}{1-cosx}|$

Como cosx≤1 podemos tirar o modulo:

$\boxed{-\frac12 ln(\frac{1+cosx}{1-cosx})+C}$

Answer 2

Resposta:

∫CosSecx dx=

∫CosSecx . CosSecx-Cotgx/CosSecx-Cotgx dx=

∫CosSec^2x-CosSecxCotgx/CosSecx-Cotgx dx =

u= CosSecx-Cotgx du= - CosSecxCotg + CosSec^2x

∫1/u . du =

ln | u | + c = l

ln | CosSecx-Cotgx | +c