tugas matematika teknik ii (integral trigonometri)
TRANSCRIPT
TUGAS LATIHAN SOAL
INTEGRAL TRIGONOMETRI
Ditujukan Untuk Memenuhi Salah Satu Tugas Mata Kuliah
Matematika Terapan II
NAMA : KAMALUDIN
NIM : 1502012
Kelas : TP1A
TEKNIK PENDINGIN DAN TATA UDARA
POLITEKNIK NEGERI INDRAMAYU
(POLINDRA)
Jalan Raya Lohbener Lama No.8 e-mail: [email protected] Kec. Lohbener Kab. Indramayu
SOAL:1. ∫cos (2 x+5)d x
2. ∫sin (3 x+2)dx
3. ∫ x3 sin 2x dx
4. ∫ x2 cos2x+5dx
PENYELESAIAN
1. ∫cos (2 x+5)dx=d (2 x+5)dx
=2
d (2x+5 )=2dx
dx=12d(2 x+5)
∫cos (2 x+5) 12d (2 x+5)
12∫ cos (2x+5)d (2x+5) = 1
2sin(2 x+5)+C
2. ∫sin (3 x+2)dx=d (3 x+2)dx
=3
d (3 x+2 )=3dx
dx=13d (3 x+2)
∫sin (3 x+2) 13d (3 x+2)
13∫sin(3 x+2)d (3x+2) = 1
3−cos (3 x+2 )+C=−1
3cos(3 x+2)+C
3. ∫ x3 sin 2x dx
= (x¿¿3).(−12
cos2x )−(3 x2 ) .(−14
sin 2x )+ (6 x ) .( 18
cos2 x )−(6 ) . ( 116
sin 2x )¿
= −12x3cos2 x+ 3
4x2 sin 2x+ 6
8cos 2x− 6
16sin2 x
= −12x3cos2 x+ 3
4x2 sin 2x+ 3
4cos2 x−3
8sin 2 x
= 34x2sin 2 x−3
8sin 2x−1
2x3 cos2 x+ 3
4cos 2x
= ( 34x2−3
8 )sin 2x – (12x3+ 3
4 )cos2 x+C
Turunan U Integral dvx3
sin 2 x
3 x2
−12
cos2 x
6 x−14
sin 2x
618
cos 2x
0 116
sin 2 x
4. ∫ x2 cos(2 x+5)dx
= (x¿¿2) .( 12
sin (2x+5))−(2x ) .(−14
cos (2x+5))+ (2 ) .(−18
sin(2 x+5))¿
= 12x2sin (2 x+5 )+ 2
4x cos (2 x+5 )−2
8sin(2x+5)
= 12x2sin (2 x+5 )+ 1
2x cos (2 x+5 )−1
4sin(2 x+5)
= 12x2sin (2 x+5 )−1
4sin(2 x+5) +1
2xcos (2x+5 )
= ( 12x2−1
4 )sin (2 x+5 )+ 12xcos (2 x+5 )+C
Turunan U Integral dvx2
cos (2x+5)
2 x12
sin(2 x+5)
2−14
cos¿)
0 −18
sin(2 x+5)