Homework week 8 solutions
Daniel Rakotonirina
March 10, 2017
2
Z
5x
dx
x2 − x − 6
1. Evaluate the following integral
−1
(2.5 marks)
Solution:
x2
5x
A1
A2
A1 (x + 2) + A2 (x − 3)
=
+
=
(1 mark)
−x−6
x−3 x+2
(x − 3)(x + 2)
We have:
5x = (A1 + A2 )x + 2A1 − 3A2
(1 mark):
From that we have a system of 2 equations
5 = A1 + A2
(1)
0 = 2A1 − 3A2
(2)
The resolution of the system gives A1 = 3 and A2 = 2. Thus,
2
2
Z 2
Z 2
Z 2
3
2
5x
dx =
dx +
dx = 3 ln |x − 3| + 2 ln |x + 2|
2−x−6
x
x
−
3
x
+
2
−1
−1
−1
−1
−1
Z
2
−1
5x
dx = − ln 4 (0.5 mark)
x2 − x − 6
1
2
Z
xe2x−1 dx (2.5 marks)
2. Evaluate the following integral
− 32
Solution:
• Integration by part
(1 mark):
Functions in original integral
u=x
dv = e2x−1 dx
Functions in new integral
du = dx
v = 12 e2x−1
Thus, the integral becomes:
Z
1
2
2x−1
xe
− 32
Z
1
2
• Substitution for
1
Z 1
x 2x−1 2
1 2 2x−1
dx = e
e
dx
3−2
2
− 23
−
2
e2x−1 dx: w = 2x − 1 =⇒ dw = 2dx
(1 mark)
− 23
Z
1
2
e
− 23
2x−1
Z
0
dx =
e
−4
1
0
1 w 1
= e = (1 − e−4 )
2
2
2
−4
w dw
Finally, we have
(0.5 mark):
Z 12
1 3
1
1
xe2x−1 dx = + e−4 − (1 − e−4 ) = + e−4
2 4
4
4
− 23
Z
3. Evaluate the following integral
ln(tan x)
dx
sin x cos x
(2.5 marks)
Solution:
• Substitution u = tan x. So, du = sec2 x dx. Thus,
Z
Z
Z
ln u
du
ln u
ln(tan x)
dx =
=
2
sin x cos x
sin x cos x sec x
sin x cos x
du
1
cos2 x
Z
=
ln u
du =
tan x
Z
ln u
du (1 mark)
u
Z
ln u
1
du using a second substitution w = ln u =⇒ dw = du
u
u
Z
Z
Z
w2
ln2 u
ln2 (tan x)
ln u
w
du =
· u dw = w dw =
+c=
+c=
+ c (1 mark)
u
u
2
2
2
• Deal with
Finally,
ln2 (tan x)
ln(tan x)
dx =
+ c (0.5 mark)
sin x cos x
2
Z
Z √
4. Evaluate the following integral
25x2 − 4
dx
x
(2.5 marks)
Solution:
We have:
Z √
q
25x2
−4
x
Z
x2 −
dx = 5
4
25
x
dx
2
2
sec θ =⇒ dx = sec θ tan θ dθ (0.5 mark)
5
5
r
r
p
4
4
4
2
2
x2 −
=
sec2 θ −
=
tan2 θ = tan θ (0.5 mark)
25
25
25
5
5
We use trigonometric substitution with x =
So,
Z √
25x2 − 4
dx = 5
x
Z
2/5 tan θ
(2/5 sec θ tan θ dθ)
2/5 sec θ
Z
Z
= 2 tan2 θ dθ = 2 (sec2 θ − 1) dθ
Z
=2
sec2 θ dθ − 2
Z
dθ
= 2 tan θ − 2θ + c (0.5 mark)
Since x =
2
sec θ, we get
5
5x
2
= sec θ so cos θ =
2
2
. Hence, θ = arccos( 5x
).
5x
(0.5 mark)
Finally,
Z √
r
25x2 − 4
4
2
dx = 2 tan θ − 2θ + c = 5 x2 −
− 2 arccos( ) + c (0.5 mark)
x
25
5x
2