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∫ sin 2 x sin 4 x dx = ....

   

  1. begin mathsize 14px style negative 3 left parenthesis sin space 2 straight x space cos space 4 straight x plus sin thin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end style  

  2. begin mathsize 14px style negative 1 third left parenthesis sin space 2 straight x space cos space 4 straight x plus sin thin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end style  

  3. begin mathsize 14px style 1 third left parenthesis sin space 2 straight x space cos space 4 straight x minus sin thin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end style     

  4. begin mathsize 14px style 3 left parenthesis sin space 2 straight x space cos space 4 straight x minus sin thin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end style   

  5. undefined  

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Jawaban terverifikasi

Jawaban

jawabannya adalah B.

jawabannya adalah B.

Pembahasan

Kita cari hasil integral di atas dengan menggunakan integral parsial. Misalkan u = sin ⁡2x dan , maka dan sehingga Selanjutnya, kita gunakan metode parsial untuk mencari hasil integral Misalkan u = cos ⁡2x dan , maka dan Sehingga Jadi, jawabannya adalah B.

Kita cari hasil integral di atas dengan menggunakan integral parsial.
Misalkan u = sin ⁡2x dan begin mathsize 14px style dv over dx equals sin space 4 straight x end style, maka

begin mathsize 14px style du over dx equals 2 space cos space 2 straight x space semicolon end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight v equals cell integral dv over dx dx end cell row blank equals cell integral sin space 4 straight x space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell negative 1 fourth cos space 4 straight x plus straight C end cell end table end style

sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell sin space 2 straight x times open parentheses negative 1 fourth cos space 4 straight x close parentheses minus integral open parentheses negative 1 fourth cos space 4 straight x close parentheses bullet left parenthesis 2 space cos space 2 straight x right parenthesis space dx end cell row blank equals cell negative 1 fourth sin space 2 straight x space cos space 4 straight x plus 1 half integral cos space 2 straight x space cos space 4 straight x space dx end cell end table end style

Selanjutnya, kita gunakan metode parsial untuk mencari hasil integralbegin mathsize 14px style integral cos space 2 straight x space cos space 4 straight x space dx end style
Misalkan u = cos ⁡2x dan begin mathsize 14px style dv over dx equals cos thin space 4 straight x end style, maka

begin mathsize 14px style du over dx equals negative 2 space sin space 2 straight x end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight v equals cell integral dv over dx dx end cell row blank equals cell integral cos space 4 straight x space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell 1 fourth sin space 4 straight x plus straight C end cell end table end style

Sehingga

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 fourth sin space 2 straight x space cos space 4 straight x plus 1 half integral cos space 2 straight x space cos space 4 straight x space dx end cell row cell integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 fourth sin space 2 straight x space cos space 4 straight x plus 1 half open parentheses open parentheses negative 2 space sin space 2 straight x close parentheses times 1 fourth sin space 4 straight x minus integral 1 fourth space sin space 4 straight x times left parenthesis negative 2 space sin space 2 straight x right parenthesis space dx close parentheses plus straight C end cell row cell integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 fourth space sin space 2 straight x space cos space 4 straight x minus 1 fourth space sin space 2 straight x space sin space 4 straight x plus 1 fourth integral sin space 2 straight x space sin space 4 straight x space dx plus straight C end cell row cell integral sin space 2 straight x space sin space 4 straight x space dx minus 1 fourth integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 fourth space sin thin space 2 straight x space cos space 4 straight x minus 1 fourth space sin space 2 straight x space sin space 4 straight x plus straight C end cell row cell 3 over 4 integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 fourth left parenthesis sin space 2 straight x space cos space 4 straight x plus sin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end cell row cell integral sin space 2 straight x space sin space 4 straight x space dx end cell equals cell negative 1 third left parenthesis sin space 2 straight x space cos space 4 straight x plus sin space 2 straight x space sin space 4 straight x right parenthesis plus straight C end cell end table end style

Jadi, jawabannya adalah B.

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