Examen 2 (solutions)(pas fini!) 201-NYB Calcul Int´ egral
14 d´ ecembre 2007 Professeur : Dimitri Zuchowski
Question 1.
a) Z
xe
xdx = xe
x− Z
e
xdx = xe
x− e
x+ C u = x dv = e
xdx du = dx v = e
xb)
Z x + 3
(x − 1)(x + 5) dx A
x − 1 + B
x + 5 = A(x + 5) + B(x − 1)
(x − 1)(x + 5) = x + 3 (x − 1)(x + 5)
=
Z 2
3(x − 1) + 1
3(x + 5) dx x = 1 ⇒ A = 2
3 et x = −5 ⇒ B = 1 3 2 ln |x − 1|
3 + ln |x + 5|
3 + C
c) Z
csc
5x cot
3x dx = Z
csc
4x cot
2x csc x cot x dx u = csc x
= Z
csc
4x(csc
2x − 1) csc x cot x dx = − Z
u
4(u
2− 1) du du = − csc x cot xdx
− u
77 + u
55 + C = − csc
7x
7 + csc
5x 5 + C
d) I =
Z
e
5xsin x dx = −e
5xcos x + 5 Z
e
5xcos x dx u = e
5xdv = sin xdx du = 5e
5xdx v = − cos x
= −e
5xcos x + 5
e
5xsin x − 5 Z
e
5xsin x dx
u = e
5xdv = cos xdx du = 5e
5xdx v = sin x
= −e
5xcos x + 5e
5xsin x − 25I
⇒ I = −e
5xcos x + 5e
5xsin x
26 + C
1
page 2 Examen 2 (solutions)(pas fini!)
e)
Z x
3√
4 − 2x
2dx =
Z 2
4sin
3θ cos θdθ 2
3cos θ
x = 2 sin θ
√ 2 dx = 2 cos θ
√ 2 dθ p 4 − 2x
2= 2 cos θ
= 2 Z
sin
3θdθ = 2 Z
(1 − cos
2θ) sin θdθ = −2 Z
(1 − u
2)du
= −2 cos θ + 2 cos
33 + C = − p
4 − 2x
2+
p (4 − 2x
2)
33 · 2
2+ C u = cos θ du = − sin θ
f) Z
x
3arctan(x
2) dx y = x
2, dy = 2xdx
= 1 2
Z
y arctan(y) dx = 1 2
y
2arctan y
2 − 1
2
Z y
21 + y
2dy
u = arctan y dv = ydy du = dy
1 + y
2v = y
22
= x
4arctan x
24 − 1
4 Z
1 − 1 1 + y
2dy
= x
4arctan x
24 − x
24 − arctan x
24 + C
g) Z
sin
3(2x) cos
2x dx = 1 2
Z
sin
3(2x)(1 + cos(2x)) dx u = cos(2x)
= 1 2
Z
sin
2(2x)(1 + cos(2x)) sin(2x) dx du = 2 sin(2x)dx
= 1 2
Z
(1 − cos
2(2x))(1 + cos(2x)) sin(2x) dx
= 1 4
Z
(1 − u
2)(1 + u) du = 1 4
Z
1 + u − u
2− u
3du cos(2x)
4 + cos
2(2x)
8 − cos
3(2x)
12 − cos
4(2x) 16 + C
Calcul Int´ egral – 201-NYB – Hiver 2007
Examen 2 (solutions)(pas fini!) page 3
h) Z
p e
2x− 4 dx = Z √
u
2− 4 u du =
Z 2
2tan θ sec θ tan θdθ 2 sec θ
u = e
xdu = e
xdx u = 2 sec θ du = 2 sec θ tan θ
√
u
2− 4 = 2 tan θ
= 2 Z
sec
2θ − 1dθ
= 2 tan θ − 2θ + C = p
e
2x− 4 − 2 arctan
√ e
2x− 4
2
! + C
i) x
3