Question

6) Calcule a integral

$\int\ \frac{dx}{\sqrt{4x^2-9} } \,$

Answer 1

Oi Lucas

∫ dx/√(4x² - 9) 

x = 3sec(u)/2 e dx = 3/2 tg(u)*sec(u) du 

√(4x² - 9) = √(9sec²(u) -9) = 3tg(u) e u = sec^(-1)((2x(3)

3/2 ∫ sec(u)/3 du = 1/2 ∫ sec(u) du = 

1/2 ∫ (sec²(u) + tg(u)*sec(u))/(tg(u) + sec(u)) du

s = tg(u) + sec(u) e ds = (sec²(u) + tg(u))/(tg(u) + sec(u)) du

1/2 ∫ 1/s ds = ln(s) + C

substitui 

∫ dx/√(4x² - 9) = 1/2 ln(√(4x² - 9) + 2x) + C 

 

Answer 2

$Ola'~~ \\ Por~substituic\~ao~trigonom\acute{e}tica~imagem ~no~ anexo \\ \\ do~anexo~temos:\begin{cases}\boxed{x= \frac{3}{2}csc \theta} ~~--\ \textgreater \ derivando \\ \boxed{dx=- \frac{3}{2}csc\theta.ctg\theta d\theta} \end{cases}\\ \\ \int \frac{1}{ \sqrt{4x\²-9} }dx~~--\ \textgreater \ substituindo \\ \\- \int \frac{1}{ \sqrt{4( \frac{3}{2}csc\theta)\²-9 } } . \frac{3}{2}csc\theta.ctg\theta d\theta \\ \\ ~~~$

$-\int \frac{1}{ \not3\sqrt{csc\²\theta-1} }. \frac{\not3}{2} csc\theta.ctg\theta d\theta ~~--\ \textgreater \ [csc\²\theta-1=ctg\²\theta] \\ \\ - \frac{1}{2} \int \frac{1}{\not ctg\theta}.csc\theta.\not ctg\theta.d\theta \\ \\ - \frac{1}{2} \int csc\theta d\theta~~--\ \textgreater \ integrando \\ \\ \frac{1}{2}ln\bigg|csc\theta+cot\theta\bigg|+C~~--\ \textgreater \ \begin{cases}csc\theta= \frac{2x}{3} \\ ctg\theta= \frac{ \sqrt{4x\²-9} }{3} \end{cases} \\ \\ \boxed{ \frac{1}{2}ln\big| \frac{2x}{3}+ \frac{ \sqrt{4x\²-9} }{3}\big|+C}$

$\\ \\ \mathbb{nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Bons~estudos!!$