Penjelasan dengan langkah-langkah:
Integral trigonometri
9. ʃ sin² 3x dx → ( 3x = u )
dx/du = ⅓
= ⅓ ʃ sin² u du
= ⅓ ʃ ½.(1-cos(2u) du
= ⅓.½ ʃ 1-cos(2u) du
= ⅙ ( u - ½sin(2u) )
= ⅙ (3x - ½sin(6x)
= ½x - 1/12 sin(6x)
= ½x - 1/12 sin(6x) + C
10. ʃ sin³ 2x dx → ( 2x = u )
dx/du = ½
= ½ ʃ sin³ u du
= ½ ʃ sin² u.sin u du
= ½ ʃ 1 - cos² u du
= ½ ( -cos u + ⅓ cos³ u )
= -½ cos 2x + ⅙ cos³ 2x
= -½ cos 2x + ⅙ cos³ 2x + C
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Verified answer
Penjelasan dengan langkah-langkah:
Integral trigonometri
9. ʃ sin² 3x dx → ( 3x = u )
dx/du = ⅓
= ⅓ ʃ sin² u du
= ⅓ ʃ ½.(1-cos(2u) du
= ⅓.½ ʃ 1-cos(2u) du
= ⅙ ( u - ½sin(2u) )
= ⅙ (3x - ½sin(6x)
= ½x - 1/12 sin(6x)
= ½x - 1/12 sin(6x) + C
10. ʃ sin³ 2x dx → ( 2x = u )
dx/du = ½
= ½ ʃ sin³ u du
= ½ ʃ sin² u.sin u du
= ½ ʃ 1 - cos² u du
= ½ ( -cos u + ⅓ cos³ u )
= -½ cos 2x + ⅙ cos³ 2x
= -½ cos 2x + ⅙ cos³ 2x + C