j) ∫ sec 5 ⁡ ( x ) d x

Passo 1

I = ∫ sec 5 ⁡ ( x ) d x

Como temos um expoente ímpar, a estratégia é utilizar a seguinte fórmula de recorrência:

∫ sec n ⁡ x   d x = sec n - 2 ⁡ ( x ) ⋅ t g ( x ) n - 1 + n - 2 n - 1 ⋅ ∫ sec n - 2 ⁡ x d x

Passo 2

Agora vamos arrumar as coisas!

I = ∫ sec 5 ⁡ ( x ) d x = sec 5 - 2 ⁡ ( x ) ⋅ t g ( x ) 5 - 1 + 5 - 2 5 - 1 ⋅ ∫ sec 5 - 2 ⁡ x d x

Fazendo as contas:

I = sec 3 ⁡ ( x ) ⋅ t g ( x ) 4 + 3 4 ⋅ ∫ sec 3 ⁡ x d x

Agora queremos integrar s e c 3 ( x ) ! Expoente ímpar, então mesma ideia.

∫ sec 3 ⁡ ( x ) d x = sec 3 - 2 ⁡ ( x ) ⋅ t g ( x ) 3 - 1 + 3 - 2 3 - 1 ⋅ ∫ sec 3 - 2 ⁡ x d x

Fazendo as contas:

∫ sec 3 ⁡ ( x ) d x = sec ⁡ ( x ) ⋅ t g ( x ) 2 + 1 2 ⋅ ∫ sec ⁡ x d x

Não esquece: ∫ sec ⁡ x d x = l n | s e c x + t g ( x ) | + c

Logo,

∫ sec 3 ⁡ ( x ) d x = sec ⁡ ( x ) ⋅ t g ( x ) 2 + 1 2 ⋅ l n | s e c x + t g ( x ) | + c 1

Passo 3

Agora vamos finalizar!

I = sec 3 ⁡ ( x ) ⋅ t g ( x ) 4 + 3 4 ⋅ ∫ sec 3 ⁡ x d x

Então nossa integral fica assim ó:

I = sec 3 ⁡ ( x ) ⋅ t g ( x ) 4 + 3 4 ⋅ sec ⁡ x ⋅ t g x 2 + 1 2 ⋅ l n s e c x + t g x + c 1

Fazendo as contas, temos:

I = sec 3 ⁡ ( x ) ⋅ t g ( x ) 4 + 3 ⋅ sec ⁡ x ⋅ t g x 8 + 3 8 ⋅ l n s e c x + t g x + c

Resposta

sec 3 ⁡ ( x ) ⋅ t g ( x ) 4 + 3 ⋅ sec ⁡ x ⋅ t g x 8 + 3 8 ⋅ l n s e c x + t g x + c

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