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Pertanyaan

Nilai dari ∫ cos x cos 2 x dx adalah ....

Nilai dari  adalah ....  

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

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

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

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

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

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Jawaban terverifikasi

Jawaban

jawabannya adalah E.

jawabannya adalah E.

Pembahasan

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

Kita cari hasil integral di atas dengan menggunakan integral parsial.
Misalkan u = cos ⁡x 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 negative sin space 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 cos space 2 straight x space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell 1 half space 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 cos space straight x space cos space 2 straight x space dx end cell equals cell cos space straight x times 1 half space sin space 2 straight x minus integral 1 half space sin space 2 straight x times left parenthesis negative sin space straight x right parenthesis space dx end cell row blank equals cell 1 half space cos space straight x space sin space 2 straight x plus 1 half integral sin space straight x space sin 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 sin space straight x space sin space 2 straight x space dx end style
Misalkan u = sin ⁡x 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 cos space 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 space 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 cos space straight x space cos space 2 straight x space dx end cell equals cell 1 half cos space straight x space sin space 2 straight x plus 1 half integral sin space straight x space sin space 2 straight x space dx end cell row cell integral cos space straight x space cos 2 straight x space dx end cell equals cell 1 half cos space straight x space sin space 2 straight x plus 1 half open parentheses sin space straight x times open parentheses negative 1 half space cos space 2 straight x close parentheses minus integral open parentheses negative 1 half space cos space 2 straight x close parentheses times cos space straight x space dx close parentheses plus straight C end cell row cell integral cos space straight x space cos space 2 straight x space dx end cell equals cell 1 half cos space straight x space sin space 2 straight x minus 1 fourth sin space straight x space cos space 2 straight x plus 1 fourth integral cos space straight x space cos space 2 straight x space dx plus straight C end cell row cell integral cos space straight x space cos space 2 straight x space dx minus 1 fourth integral cos space straight x space cos space 2 straight x space dx end cell equals cell 1 half space cos space straight x space sin space 2 straight x minus 1 fourth space sin space straight x space cos space 2 straight x plus straight C end cell row cell 3 over 4 integral cos space straight x space cos space 2 straight x space dx end cell equals cell 1 fourth left parenthesis 2 space cos space straight x space sin space 2 straight x minus sin space straight x space cos space 2 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 1 third left parenthesis 2 space cos space straight x space sin space 2 straight x minus sin space straight x space cos space 2 straight x right parenthesis plus straight C end cell end table end style

Jadi, jawabannya adalah E.

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