Question

Calcule o valor das integrais
dica: use a formula: ∫$\frac{1}{\sqrt{a^{2} +x^{2} } } dx=ln |x+\sqrt{a^{2} +x^{2} |} +k$

Answer 1

a)

∫ dx/√(x²+9/25)

Substituindo

x= 3tan(u)/5  ==>dx=3sec²(u)/5 du

∫ 3sec²(u)/5 du/√([(3tan(u)/5)²+9/25]

∫ 3sec²(u)/5 du/√[9tan²(u)/25+9/25]

∫ 3sec²(u)/5 du/√[9sen²(u)/25+9cos²(u)]/25cos²(u)]

∫ 3sec²(u)/5 du/√[9/25cos²(u)]

∫ 3sec²(u)/5 du/[3*sec(u)/5]

∫ sec(u) du

∫ sec(u) * ( tan(u)+sec(u))/( tan(u)+sec(u)) du

Faça s=(tan(u)+sec(u)  ==> ds =[sec²(u) +tan(u)*sec(u)] du

∫ sec(u) * ( tan(u)+sec(u))/s    ds/[sec²(u) +tan(u)*sec(u)]

∫( tan(u)*sec(u)+sec²(u))/s    ds/[sec²(u) +tan(u)*sec(u)]

∫1 /s    ds

= ln |s|+ c

Como s= (tan(u)+sec(u)

= ln |(tan(u)+sec(u)|+ c

Como x= 3tan(u)/5 ==> tan(u) =5x/3 ==> u = arctan(5x/3)

= ln | (tan( arctan(5x/3)  )+sec( arctan(5x/3) ) |+ c

b)

Pode ser feita da mesma maneira

Answer 2

Resposta:

a)

∫ dx/√(x²+9/25)

Substituindo

x= 3tan(u)/5  ==>dx=3sec²(u)/5 du

∫ 3sec²(u)/5 du/√([(3tan(u)/5)²+9/25]

∫ 3sec²(u)/5 du/√[9tan²(u)/25+9/25]

∫ 3sec²(u)/5 du/√[9sen²(u)/25+9cos²(u)]/25cos²(u)]

∫ 3sec²(u)/5 du/√[9/25cos²(u)]

∫ 3sec²(u)/5 du/[3*sec(u)/5]

∫ sec(u) du

∫ sec(u) * ( tan(u)+sec(u))/( tan(u)+sec(u)) du

Faça s=(tan(u)+sec(u)  ==> ds =[sec²(u) +tan(u)*sec(u)] du

∫ sec(u) * ( tan(u)+sec(u))/s    ds/[sec²(u) +tan(u)*sec(u)]

∫( tan(u)*sec(u)+sec²(u))/s    ds/[sec²(u) +tan(u)*sec(u)]

∫1 /s    ds

= ln |s|+ c

Como s= (tan(u)+sec(u)

= ln |(tan(u)+sec(u)|+ c

Como x= 3tan(u)/5 ==> tan(u) =5x/3 ==> u = arctan(5x/3)

= ln | (tan( arctan(5x/3)  )+sec( arctan(5x/3) ) |+ c

b)

Pode ser feita da mesma maneira