Iklan

Iklan

Pertanyaan

∫ − 2 π ​ 2 π ​ ​ ( sin 2 x − 2 ) ⋅ cos x d x = ...

 

Iklan

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

.

 integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript open parentheses sin squared open parentheses x close parentheses minus 2 close parentheses cos open parentheses x close parentheses straight d x equals negative 10 over 3.

Iklan

Pembahasan

Ingat bahwa: Sehingga, Misalkan: Sehingga, Jadi, .

Ingat bahwa:

  • table attributes columnalign right center left columnspacing 0px end attributes row cell integral f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses d x end cell equals cell integral f open parentheses x close parentheses d x plus-or-minus integral g open parentheses x close parentheses d x end cell row cell integral x to the power of a straight d x end cell equals cell fraction numerator x to the power of a plus 1 end exponent over denominator a plus 1 end fraction comma space a not equal to negative 1 end cell row cell integral a f left parenthesis x right parenthesis straight d x end cell equals cell a integral f left parenthesis x right parenthesis straight d x end cell row cell integral cos space left parenthesis x right parenthesis straight d x end cell equals cell sin space x end cell end table 

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript open parentheses sin squared space x minus 2 close parentheses times cos space x space d x end cell equals cell integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript cos open parentheses x close parentheses sin squared open parentheses x close parentheses minus cos open parentheses x close parentheses times 2 space d x end cell row blank equals cell integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript open parentheses cos open parentheses x close parentheses sin squared open parentheses x close parentheses minus 2 cos open parentheses x close parentheses close parentheses space d x end cell row blank equals cell integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript cos open parentheses x close parentheses sin squared open parentheses x close parentheses d x minus integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript 2 cos open parentheses x close parentheses d x end cell end table 

Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell sin open parentheses x close parentheses end cell row cell straight d u end cell equals cell cos open parentheses x close parentheses straight d x end cell row cell straight d x end cell equals cell fraction numerator 1 over denominator cos open parentheses x close parentheses end fraction straight d u end cell row blank blank blank row u equals cell sin open parentheses x close parentheses end cell row x equals cell negative pi over 2 rightwards arrow u equals negative 1 end cell row x equals cell pi over 2 rightwards arrow u equals 1 end cell end table 

Sehingga,

integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript cos open parentheses x close parentheses sin squared open parentheses x close parentheses straight d x minus integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript 2 cos open parentheses x close parentheses straight d x equals integral subscript negative 1 end subscript superscript 1 u squared straight d u minus 2 times integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript cos open parentheses x close parentheses straight d x equals open square brackets u cubed over 3 close square brackets subscript negative 1 end subscript superscript 1 minus 2 open square brackets sin open parentheses x close parentheses close square brackets subscript negative pi over 2 end subscript superscript pi over 2 end superscript equals open square brackets 1 third minus open parentheses negative 1 third close parentheses close square brackets minus open square brackets 2 times thin space 2 close square brackets equals 2 over 3 minus 4 equals negative 10 over 3 

Jadi, integral subscript negative pi over 2 end subscript superscript pi over 2 end superscript open parentheses sin squared open parentheses x close parentheses minus 2 close parentheses cos open parentheses x close parentheses straight d x equals negative 10 over 3.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

2

Nayyara Zahra

Pembahasan tidak menjawab soal Jawaban tidak sesuai Pembahasan lengkap banget

Iklan

Iklan

Pertanyaan serupa

∫ sin 3 x . cos x d x = ...

28

4.4

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia