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∫ e x sin 2 x dx = ....

  

  1. begin mathsize 14px style 1 fifth straight e to the power of straight x left parenthesis 2 cos space 2 straight x minus sin space 2 straight x right parenthesis plus straight C end style

  2. begin mathsize 14px style negative 5 straight e to the power of straight x left parenthesis 2 cos space 2 straight x minus sin space 2 straight x right parenthesis plus straight C end style   

  3. begin mathsize 14px style negative 1 fifth straight e to the power of straight x left parenthesis 2 cos space 2 straight x minus sin space 2 straight x right parenthesis plus straight C end style   

  4. begin mathsize 14px style 1 fifth straight e to the power of straight x left parenthesis 2 cos space 2 straight x plus sin space 2 straight x right parenthesis plus straight C end style   

  5. begin mathsize 14px style negative 1 fifth straight e to the power of straight x left parenthesis 2 cos space 2 straight x plus sin space 2 straight x right parenthesis plus straight C end style

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R. RGFLLIMA

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Jawaban terverifikasi

Jawaban

jawabannya adalah C.

jawabannya adalah C.

Pembahasan

Kita cari hasil integral di atas dengan menggunakan integral parsial. Misalkan dan , maka dan sehingga Selanjutnya, kita gunakan metode parsial untuk mencari hasil integral Misalkan dan , maka dan Sehingga Jadi, jawabannya adalah C.

Kita cari hasil integral di atas dengan menggunakan integral parsial.
Misalkan begin mathsize 14px style straight u equals straight e to the power of straight x end style dan begin mathsize 14px style dv over dx equals sin space 2 straight x end style, maka

begin mathsize 14px style du over dx equals straight e to the power of straight x 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 2 straight x space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell negative 1 half space cos space 2 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 straight e to the power of straight x sin space 2 straight x space dx end cell equals cell straight e to the power of straight x times open parentheses negative 1 half cos space 2 straight x close parentheses minus integral open parentheses negative 1 half cos space 2 straight x close parentheses times straight e to the power of straight x space dx end cell row blank equals cell negative 1 half space straight e to the power of straight x space cos space 2 straight x plus 1 half integral straight e to the power of straight x space cos space 2 straight x space dx end cell end table end style

Selanjutnya, kita gunakan metode parsial untuk mencari hasil integral begin mathsize 14px style integral straight e to the power of straight x space cos space 2 straight x space dx end style
Misalkan begin mathsize 14px style straight u equals straight e to the power of straight x end style dan begin mathsize 14px style dv over dx equals cos space 2 straight x end style, maka

begin mathsize 14px style du over dx equals straight e to the power of straight x 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 cos space 2 straight x space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell 1 half sin space 2 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 straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 half straight e to the power of straight x space cos space 2 straight x plus 1 half integral straight e to the power of straight x space cos space 2 straight x space dx end cell row cell integral straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 half straight e to the power of straight x space cos space 2 straight x plus 1 half open parentheses straight e to the power of straight x times 1 half sin space 2 straight x minus integral 1 half space sin space 2 straight x times straight e to the power of straight x space dx close parentheses plus straight C end cell row cell integral straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 half straight e to the power of straight x space cos space 2 straight x plus 1 fourth straight e to the power of straight x space sin space 2 straight x minus 1 fourth integral straight e to the power of straight x space sin space 2 straight x space dx plus straight C end cell row cell integral straight e to the power of straight x space sin space 2 straight x space dx plus 1 fourth integral straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 half straight e to the power of straight x space cos space 2 straight x plus 1 fourth straight e to the power of straight x space end exponent sin space 2 straight x plus straight C end cell row cell 5 over 4 integral straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 fourth straight e to the power of straight x open parentheses 2 cos space 2 straight x minus sin space 2 straight x close parentheses plus straight C end cell row cell integral straight e to the power of straight x space sin space 2 straight x space dx end cell equals cell negative 1 fifth straight e to the power of straight x left parenthesis 2 cos space 2 straight x minus sin space 2 straight x right parenthesis plus straight C end cell end table end style

Jadi, jawabannya adalah C.

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