Iklan

Pertanyaan

Nilai dari ∫ sin − 1 2 x d x adalah ....

Nilai dari  adalah ....

  1. begin mathsize 14px style x space sin to the power of negative 1 end exponent space 2 x minus 1 fourth square root of 1 minus 4 x squared end root plus C end style

  2. begin mathsize 14px style sin to the power of negative 1 end exponent space 2 x minus 1 fourth square root of 1 minus 4 x squared end root plus C end style

  3. begin mathsize 14px style sin to the power of negative 1 end exponent space 2 x minus 1 half square root of 1 minus 4 x squared end root plus C end style

  4. begin mathsize 14px style x space sin to the power of negative 1 end exponent space 2 x minus 1 half square root of 1 minus 4 x squared end root plus C end style

  5. begin mathsize 14px style x space sin to the power of negative 1 end exponent space 2 x plus 1 half square root of 1 minus 4 x squared end root plus C end style

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

01

:

15

:

22

:

40

Klaim

Iklan

F. Freelancer10

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

Pembahasan

Kita cari hasil integral di atas dengan menggunakan integral parsial. Misalkan dan , maka dan Oleh karena itu Gunakan metode substitusi untuk mencari hasil integral . Misalkan , maka Dengan demikian Oleh karena itu Jadi, jawaban yang tepat adalah E.

Kita cari hasil integral di atas dengan menggunakan integral parsial.
Misalkan begin mathsize 14px style u equals sin to the power of negative 1 end exponent space 2 x end style dan begin mathsize 14px style fraction numerator straight d v over denominator straight d x end fraction equals 1 end style, maka

begin mathsize 14px style fraction numerator straight d u over denominator straight d x end fraction equals fraction numerator 2 over denominator square root of 1 minus 4 x squared end root end fraction semicolon space ingat space kembali space turunan space dari space sin to the power of negative 1 end exponent space 2 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 1 space dx semicolon end cell row blank equals cell straight x plus straight C end cell end table end style

Oleh karena itu

begin mathsize 14px style integral sin to the power of negative 1 end exponent space 2 straight x space dx equals sin to the power of negative 1 end exponent space 2 straight x times straight x minus integral straight x times fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction dx end style

Gunakan metode substitusi untuk mencari hasil integral begin mathsize 14px style integral straight x times fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction dx end style.
Misalkan begin mathsize 14px style straight u equals 1 minus 4 straight x squared end style, maka

begin mathsize 14px style du over dx equals negative 8 straight x end style

Dengan demikian

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral straight x times fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction dx end cell equals cell negative 1 fourth integral fraction numerator negative 8 straight x over denominator square root of 1 minus 4 straight x squared end root end fraction dx end cell row blank equals cell negative 1 fourth integral fraction numerator 1 over denominator square root of straight u end fraction du over dx dx end cell row blank equals cell negative 1 half square root of straight u plus straight C end cell row blank equals cell negative 1 half square root of 1 minus 4 straight x squared end root plus straight C end cell end table end style

Oleh karena itu

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral sin to the power of negative 1 end exponent space 2 straight x space dx end cell equals cell sin to the power of negative 1 end exponent space 2 straight x times straight x minus integral straight x times fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction dx end cell row blank equals cell straight x space sin to the power of negative 1 end exponent space 2 straight x plus 1 half square root of 1 minus 4 straight x squared end root plus straight C end cell end table end style

Jadi, jawaban yang tepat adalah E.

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

4

Iklan

Pertanyaan serupa

∫ cos − 1 2 x dx = ....

1

0.0

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