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1. Diketahui (3x2 + 2x + 1) dx = 25 Nilai a = … 
A. – 4 
B. – 2 
C. – 1 
D. 1 
E. 2 
PEMBAHASAN : 
(3x2 + 2x + 1) dx = x3 + x2 + x 
25 = (33 + 32 + 3) – (a3 + a2 + a) 
a3 + a2 + a = 27 + 9 + 3 – 25 
a3 + a2 + a – 14 = 0 
(a – 2)(a2 + a + 7) = 0 
a = 2 atau a2 + a + 7 = 0 
jadi a = 1 
JAWABAN : D 
2. Nilai sin 2x cos x dx = … 
A. -4/3 
B. -1/3 
C. 1/3 
D. 2/3 
E. 4/3 
PEMBAHASAN : 
sin 2x cos x dx = 2 sin x cos x cos x dx
= 2 sin x cos2 x dx 
misal u = cos x du = -sin x dx 
= 2 u2 (-du) 
= - u3 
Substitusi u = cos x 
= - cos3 x 
= - cos3 + cos3 0 
= - (-1)3 + .13 
= + 
= 
JAWABAN : D 
3. Hasil dari 3x dx = … 
A. 7/2 
B. 8/3 
C. 7/3 
D. 4/3 
E. 2/3 
PEMBAHASAN : 
3x dx = … 
misal u = 3x2 + 1 du = 6x dx 
=
= u1/2 du 
= . u3/2 
substitusi u = 3x2 + 1, sehingga diperoleh 
= (3x2 + 1)3/2 
= (3.12 + 1)3/2 – (3.02 + 1)3/2 
= 8 – .1 
= 
JAWABAN : C 
4. Hasil dari cos5 x dx = … 
A. - cos6 x sin x + C 
B. cos6 x sin x + C 
C. –sin x + sin3 x + sin5 x + C 
D. sin x – sin3 x + sin5 x + C 
E. sin x + sin3 x + sin5 x + C 
PEMBAHASAN : 
cos5 x dx = cos x (cos2 x)2 dx 
= cos x (1 – sin2 x)2 dx 
= cos x (1 – 2 sin2 x + sin4 x) dx 
misal u = sin x du = cos x 
= (1 – 2u2 + u4) du
= u – u3 + u5 + C 
substitusi u = sin x, 
= sin x – sin3 x + sin5 x + C 
JAWABAN : D 
5. Hasil dari cos x (x2 + 1) dx = … 
A. x2 sin x + 2x cos x + C 
B. (x2 – 1)sin x + 2x cos x + C 
C. (x2 + 3)sin x – 2x cos x + C 
D. 2x2 cos x + 2x2 sin x + C 
E. 2x sin x – (x2 – 1)cos x + C 
PEMBAHASAN : 
dalam penyelesaian soal ini akan menggunakan Integral Parsial 
u = x2 + 1 du = 2x dx 
dv = cos x dx v = sin x 
u dv = uv – v du 
= sin x (x2 + 1) – sin x 2x dx 
parsial lagi 
m = 2x dm = 2 dx 
dn = sin x dx n = -cos x 
= sin x (x2 + 1) – (2x (-cos x) – -cos x 2 dx) 
= sin x (x2 + 1) – (-2x cos x + 2 sin x) + C 
= sin x (x2 + 1) + 2x cos x – 2 sin x + C 
= sin x (x2 – 1) + 2x cos x + C
JAWABAN : B 
6. Diketahui (3x2 – 2x + 2) dx = 40. Nilai p = … 
A. 2 
B. 1 
C. – 1 
D. – 2 
E. – 4 
PEMBAHASAN : 
(3x2 – 2x + 2) dx = x3 – x2 + 2x 
40 = (33 – 32 + 6) – (p3 – p2 + 2p) 
p3 – p2 + 2p = 27 – 9 + 6 – 40 
p3 – p2 + 2p + 16 = 0 
(p + 2)(p2 + p + 7) = 0 
p = -2 atau p2 + p + 7 = 0 
jadi p = -1 
JAWABAN : C 
7. Hasil dari sin 3x cos 5x dx = … 
A. -10/6 
B. -8/10 
C. -5/16 
D. -4/16 
E. 0 
PEMBAHASAN :
sin 3x cos 5x dx = [sin 8x + sin (-2x)] dx 
[Sifat Trigonometri] 
= sin 8x dx – sin 2x dx 
misal u = 8x du = 8 dx 
v = 2x dv = 2 dx 
= sin u – sin v 
= - cos u + cos v 
substitusi u = 8x dan v = 2x 
= - cos 8x + cos 2x 
= [- (cos 8( ) - cos 8(0))] + [ (cos 2( ) – cos 2(0))] 
= [- (1 – 1)] + [ (-1 – 1)] 
= 
JAWABAN : 
8. x sin x dx = … 
A. 
B. 
C. 
D. 
E. 
PEMBAHASAN :
dalam penyelesaian soal ini akan menggunakan Integral Parsial 
u = x du = dx 
dv = sin x dx v = -cos x 
u dv = uv – v du 
= -x cos x – (-cos x) dx 
= [-x cos x + sin x] 
= [- cos ( ) + sin ( )] – [-0 cos 0 + sin 0] 
= - (-1) 
= 
JAWABAN : D 
9. Nilai (2x + sin x) dx = … 
A. – 1 
B. 
C. + 1 
D. – 1 
E. + 1 
PEMBAHASAN : 
(2x + sin x) dx = x2 – cos x 
= (( )2 – cos ( )) – (02 – cos 0) 
= ( – 0) – (02 – 1) 
= + 1 
JAWABAN : C
10. Nilai x sin(x2 + 1) dx = … 
A. –cos (x2 + 1) + C 
B. cos (x2 + 1) + C 
C. –½ cos (x2 + 1) + C 
D. ½ cos (x2 + 1) + C 
E. –2cos (x2 + 1) + C 
PEMBAHASAN : 
misal u = x2 + 1 du = 2x dx 
x sin(x2 + 1) dx = sin u 
= cos u + C 
substitusi u = x2 + 1 
= cos (x2 + 1) + C 
JAWABAN : C

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Integral

  • 1. 1. Diketahui (3x2 + 2x + 1) dx = 25 Nilai a = … A. – 4 B. – 2 C. – 1 D. 1 E. 2 PEMBAHASAN : (3x2 + 2x + 1) dx = x3 + x2 + x 25 = (33 + 32 + 3) – (a3 + a2 + a) a3 + a2 + a = 27 + 9 + 3 – 25 a3 + a2 + a – 14 = 0 (a – 2)(a2 + a + 7) = 0 a = 2 atau a2 + a + 7 = 0 jadi a = 1 JAWABAN : D 2. Nilai sin 2x cos x dx = … A. -4/3 B. -1/3 C. 1/3 D. 2/3 E. 4/3 PEMBAHASAN : sin 2x cos x dx = 2 sin x cos x cos x dx
  • 2. = 2 sin x cos2 x dx misal u = cos x du = -sin x dx = 2 u2 (-du) = - u3 Substitusi u = cos x = - cos3 x = - cos3 + cos3 0 = - (-1)3 + .13 = + = JAWABAN : D 3. Hasil dari 3x dx = … A. 7/2 B. 8/3 C. 7/3 D. 4/3 E. 2/3 PEMBAHASAN : 3x dx = … misal u = 3x2 + 1 du = 6x dx =
  • 3. = u1/2 du = . u3/2 substitusi u = 3x2 + 1, sehingga diperoleh = (3x2 + 1)3/2 = (3.12 + 1)3/2 – (3.02 + 1)3/2 = 8 – .1 = JAWABAN : C 4. Hasil dari cos5 x dx = … A. - cos6 x sin x + C B. cos6 x sin x + C C. –sin x + sin3 x + sin5 x + C D. sin x – sin3 x + sin5 x + C E. sin x + sin3 x + sin5 x + C PEMBAHASAN : cos5 x dx = cos x (cos2 x)2 dx = cos x (1 – sin2 x)2 dx = cos x (1 – 2 sin2 x + sin4 x) dx misal u = sin x du = cos x = (1 – 2u2 + u4) du
  • 4. = u – u3 + u5 + C substitusi u = sin x, = sin x – sin3 x + sin5 x + C JAWABAN : D 5. Hasil dari cos x (x2 + 1) dx = … A. x2 sin x + 2x cos x + C B. (x2 – 1)sin x + 2x cos x + C C. (x2 + 3)sin x – 2x cos x + C D. 2x2 cos x + 2x2 sin x + C E. 2x sin x – (x2 – 1)cos x + C PEMBAHASAN : dalam penyelesaian soal ini akan menggunakan Integral Parsial u = x2 + 1 du = 2x dx dv = cos x dx v = sin x u dv = uv – v du = sin x (x2 + 1) – sin x 2x dx parsial lagi m = 2x dm = 2 dx dn = sin x dx n = -cos x = sin x (x2 + 1) – (2x (-cos x) – -cos x 2 dx) = sin x (x2 + 1) – (-2x cos x + 2 sin x) + C = sin x (x2 + 1) + 2x cos x – 2 sin x + C = sin x (x2 – 1) + 2x cos x + C
  • 5. JAWABAN : B 6. Diketahui (3x2 – 2x + 2) dx = 40. Nilai p = … A. 2 B. 1 C. – 1 D. – 2 E. – 4 PEMBAHASAN : (3x2 – 2x + 2) dx = x3 – x2 + 2x 40 = (33 – 32 + 6) – (p3 – p2 + 2p) p3 – p2 + 2p = 27 – 9 + 6 – 40 p3 – p2 + 2p + 16 = 0 (p + 2)(p2 + p + 7) = 0 p = -2 atau p2 + p + 7 = 0 jadi p = -1 JAWABAN : C 7. Hasil dari sin 3x cos 5x dx = … A. -10/6 B. -8/10 C. -5/16 D. -4/16 E. 0 PEMBAHASAN :
  • 6. sin 3x cos 5x dx = [sin 8x + sin (-2x)] dx [Sifat Trigonometri] = sin 8x dx – sin 2x dx misal u = 8x du = 8 dx v = 2x dv = 2 dx = sin u – sin v = - cos u + cos v substitusi u = 8x dan v = 2x = - cos 8x + cos 2x = [- (cos 8( ) - cos 8(0))] + [ (cos 2( ) – cos 2(0))] = [- (1 – 1)] + [ (-1 – 1)] = JAWABAN : 8. x sin x dx = … A. B. C. D. E. PEMBAHASAN :
  • 7. dalam penyelesaian soal ini akan menggunakan Integral Parsial u = x du = dx dv = sin x dx v = -cos x u dv = uv – v du = -x cos x – (-cos x) dx = [-x cos x + sin x] = [- cos ( ) + sin ( )] – [-0 cos 0 + sin 0] = - (-1) = JAWABAN : D 9. Nilai (2x + sin x) dx = … A. – 1 B. C. + 1 D. – 1 E. + 1 PEMBAHASAN : (2x + sin x) dx = x2 – cos x = (( )2 – cos ( )) – (02 – cos 0) = ( – 0) – (02 – 1) = + 1 JAWABAN : C
  • 8. 10. Nilai x sin(x2 + 1) dx = … A. –cos (x2 + 1) + C B. cos (x2 + 1) + C C. –½ cos (x2 + 1) + C D. ½ cos (x2 + 1) + C E. –2cos (x2 + 1) + C PEMBAHASAN : misal u = x2 + 1 du = 2x dx x sin(x2 + 1) dx = sin u = cos u + C substitusi u = x2 + 1 = cos (x2 + 1) + C JAWABAN : C