Pertanyaan

Hasil dari adalah ....

Hasil dari $integral open parentheses fraction numerator 1 plus sec x over denominator 1 minus sec x end fraction close parentheses space d x$ adalah ....

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Habis dalam

Klaim

S. Difhayanti

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Prof. Dr. Hamka

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Gunakan beberapa identitas trigonometri berikut ini untuk menyederhanakan soal di atas! Oleh karena itu, diperoleh hasil perhitungan sebagai berikut. Jadi, jawaban yang tepat adalah B.

Gunakan beberapa identitas trigonometri berikut ini untuk menyederhanakan soal di atas!

table attributes columnalign right center left columnspacing 0px end attributes row cell cot squared x end cell equals cell csc squared x minus 1 end cell row cell tan squared x plus 1 end cell equals cell sec squared x end cell end table

Oleh karena itu, diperoleh hasil perhitungan sebagai berikut.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses fraction numerator 1 plus sec x over denominator 1 minus sec x end fraction close parentheses space d x end cell row blank equals cell integral open parentheses fraction numerator 1 plus sec x over denominator 1 minus sec x end fraction times fraction numerator 1 plus sec x over denominator 1 plus sec x end fraction close parentheses space d x end cell row blank equals cell integral fraction numerator open parentheses 1 plus sec x close parentheses squared over denominator 1 minus sec squared x end fraction space d x end cell row blank equals cell integral fraction numerator 1 plus 2 sec x plus sec squared x over denominator negative tan squared x end fraction space d x end cell row blank equals cell negative integral open parentheses fraction numerator 1 over denominator tan squared x end fraction plus fraction numerator 2 sec x over denominator tan squared x end fraction plus fraction numerator sec squared x over denominator tan squared x end fraction close parentheses space d x end cell row blank equals cell negative integral open parentheses cot squared x plus fraction numerator fraction numerator 2 over denominator cos x end fraction over denominator fraction numerator sin squared x over denominator cos squared x end fraction end fraction plus fraction numerator fraction numerator 1 over denominator cos squared x end fraction over denominator fraction numerator sin squared x over denominator cos squared x end fraction end fraction close parentheses space d x end cell row blank equals cell negative integral open parentheses open parentheses csc squared x minus 1 close parentheses plus fraction numerator 2 over denominator cos x end fraction times fraction numerator cos squared x over denominator sin squared x end fraction close end cell row blank blank cell plus open fraction numerator 1 over denominator cos squared x end fraction times fraction numerator cos squared x over denominator sin squared x end fraction close parentheses space d x end cell row blank equals cell negative integral open parentheses open parentheses csc squared x minus 1 close parentheses plus fraction numerator 2 cos x over denominator sin squared x end fraction close end cell row blank blank cell open plus fraction numerator 1 over denominator sin squared x end fraction close parentheses space d x end cell row blank equals cell negative integral open parentheses open parentheses csc squared x minus 1 close parentheses plus 2 times fraction numerator cos x over denominator sin x end fraction times fraction numerator 1 over denominator sin x end fraction close end cell row blank blank cell open plus fraction numerator 1 over denominator sin squared x end fraction close parentheses space d x end cell row blank equals cell negative integral open parentheses csc squared x minus 1 plus 2 times cot x times csc x plus csc squared x close parentheses space d x end cell row blank equals cell negative integral open parentheses 2 csc squared x plus 2 cot x csc x minus 1 close parentheses space d x end cell row blank equals cell negative open parentheses negative 2 cot x minus 2 csc x minus x close parentheses plus C end cell row blank equals cell 2 cot x plus 2 csc x plus x plus C end cell end table end style$

Jadi, jawaban yang tepat adalah B.