Março 7, 2020

por Dore

How do you integrate $csc x$ $cscx$ ?

Responda:

$\int csc x d x = - ln | csc (x) + cot (x) | + C$ $int csc x dx = - ln|csc(x) + cot(x)| +C$

Explicação:

There are many ways to prove this result. The quickest method that I am aware of is as follows:

$\int csc x d x = \int csc x \frac{csc x + cot x}{csc x + cot x} d x$ $int csc x dx = int cscx (cscx + cotx)/(cscx + cotx) dx$
$= \int \frac{{csc}^{2} x + csc x cot x}{csc x + cot x} d x$ $" "= int (csc^2x + cscxcotx)/(cscx + cotx) dx$

Then we perform simple substitution, Let

$u = csc x + cot x \Rightarrow \frac{d u}{d x} = - csc x cot x - {csc}^{2} x$ $u = cscx + cotx => (du)/dx = -cscxcotx - csc^2x$
$= - (csc x cot x + {csc}^{2} x)$ $" "= -(cscxcotx + csc^2x)$

E entao:

$\int csc x d x = \int (- \frac{1}{u}) d u$ $int csc x dx = int (-1/u) du$
$= - \int \frac{1}{u} d u$ $" "= - int 1/u du$
$= - ln | u | + C$ $" "= - ln|u| +C$
$= - ln | csc x + cot x | + C$ $" "= - ln|cscx + cotx| +C$