Pertanyaan

Hasil dari ....

Hasil dari $begin mathsize 14px style integral open parentheses fraction numerator sec squared invisible function application 3 x over denominator open parentheses 1 plus tan invisible function application 3 x close parentheses cubed end fraction close parentheses d x equals end style$ ....

$begin mathsize 14px style negative fraction numerator 1 over denominator 6 open parentheses 1 plus tan invisible function application 3 x close parentheses squared end fraction plus straight C end style$

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

.

$begin mathsize 14px style integral open parentheses fraction numerator sec squared invisible function application 3 x over denominator open parentheses 1 plus tan invisible function application 3 x close parentheses cubed end fraction close parentheses d x equals negative fraction numerator 1 over denominator 6 open parentheses 1 plus tan invisible function application 3 x close parentheses squared end fraction plus straight C end style$ .

Pembahasan

Misalkan: u = 1 + tan ⁡3x Maka, Sehingga integral di atas menjadi, Jadi, .

Misalkan:
u = 1 + tan ⁡3x

Maka,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d u over denominator d x end fraction end cell equals cell 3 sec squared invisible function application 3 x end cell row cell integral fraction numerator d u over denominator d x end fraction d x end cell equals cell 3 integral sec squared invisible function application 3 x d x end cell row cell 1 third integral d u end cell equals cell integral sec squared invisible function application 3 x d x end cell end table end style$

Sehingga integral di atas menjadi,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses fraction numerator sec squared invisible function application 3 x over denominator open parentheses 1 plus tan invisible function application 3 x close parentheses cubed end fraction close parentheses d x end cell equals cell 1 third integral 1 over u cubed d u end cell row blank equals cell 1 third integral u to the power of negative 3 end exponent d u end cell row blank equals cell 1 third times open parentheses fraction numerator 1 over denominator negative 2 end fraction u to the power of negative 2 end exponent close parentheses plus straight C end cell row blank equals cell negative fraction numerator 1 over denominator 6 open parentheses 1 plus tan invisible function application 3 x close parentheses squared end fraction plus straight C end cell end table end style$

Jadi, $begin mathsize 14px style integral open parentheses fraction numerator sec squared invisible function application 3 x over denominator open parentheses 1 plus tan invisible function application 3 x close parentheses cubed end fraction close parentheses d x equals negative fraction numerator 1 over denominator 6 open parentheses 1 plus tan invisible function application 3 x close parentheses squared end fraction plus straight C end style$ .