Question

Calcule:
1) $\int\limits {9tg(x)ln(5cos(x))} \, dx$

2) $\int\limits {\sqrt{-x^2+8x+2} } \, dx$

Answer 1

Resposta:

1) ∫ 9*tan(x) * log(5*cos(x)) dx

faça u =ln(5cos(x)) ==>du=-tan(x) dx

=9 *∫*tan(x) *u * du/(-tan(x))

= -9 ∫u du

= -9 u²/2 +c

Como u =ln(5cos(x)) , ficamos com:

=-(9/2) * ln²(5cos(x))   + c

2)  ∫ √(-x^2+8*x+2) dx

## completando os quadrados

= ∫ √(18-(x-4)²) dx

faça u=x-4 ==>du=dx

= ∫ √(18-u²) du

= ∫ √((3√2)²-u²) du

Substitua  sen(s)=u/(2√3)    ==>u= sen(s) * (3√2)

du=  (3√2)  * cos(s) ds

= ∫ √(18-18*sen²(s)) * (3√2)  * cos(s) ds

= ∫ √(18cos²(s)) * (3√2)  * cos(s) ds

= ∫ (3√2) *cos(s) * (3√2)  * cos(s) ds

=18 ∫ cos²(s) ds

#######################

Da tabela

∫ cos²(s) ds = s/2 +(1/4)* sen(2s) + c

########################

sabemos que  u=(3√2)*sen(s)

==>sen(s)=u/(3√2)

==> s =arc seno (u/(3√2)

∫ √(-x^2+8*x+2) dx = (arc seno (u/(3√2))/2 +(1/4)* sen(2* arc seno (u/(3√2)) + c

Sabemos que  u=(x-4)

∫ √(-x^2+8*x+2) dx

= (arc seno ((x-4)/(3√2))/2 +(1/4)* sen(2* arc seno ((x-4)/(3√2)) + c