Total: 30 pts (=15% of the final grade )
Time allowed: 50 minutes.
You are not allowed to use any electronic devices, such as calculators, laptops or phones, during the
test. Please show your steps clearly.
1. (10 pts) Evaluate the integral.
Z
(a) (2 pts) sin x cos x dx.
Z
(b) (3 pts) tan2 (2x) dx.
Z
(c) (5 pts)
1
tan−1 x dx.
0
Sol. (a)
Z
Z
sin x cos xdx =
Z
0
sin x(sin x) dx =
udu
(u = sin x)
u2
= +C
2
sin2 x
+ C.
=
2
(b)
Z
Z
du
2
du
tan u
)
(u = 2x ∴ dx =
2
Z
1
=
sec2 u − 1 du
(tan2 u = sec2 u − 1)
2
u
1
= tan u − + C
2
2
tan 2x
=
− x + C.
2
2
tan (2x)dx =
2
(c)
Z
1
−1
tan
0
−1
1
Z
1
x(tan−1 x)0 dx
0
π
Z 1 x
= 1· −0·0 −
dx
2
4
0 1+x
Z
π 1 1 (1 + x2 )0
= −
dx
4 2 0 1 + x2
Z
π 1 2 du
= −
dx
(u = x2 + 1)
4 2 1 u
π 1
= − [ln |u|]21
4 2
π 1
= − ln 2.
4 2
xdx = x tan
1
x
0
−
2. (10 pts)
(a) (3 pts) Compute
d
dx
sec x and express it in terms of sec x and tan x.
Z
(b) (2 pts) Prove that sec x dx = ln | sec x + tan x| + C.
Z
(c) (5 pts) Compute sec3 xdx.
d
Sol. (a) dx
(sec x) = sec x tan x. (You can write the answer directly, or derive it using the derivative
of cos x.)
(b)
d
(sec x + tan x)0
sec x tan x + sec2 x
(ln | sec x + tan x|) =
=
= sec x.
dx
sec x + tan x
sec x + tan x
So
R
sec xdx = ln | sec x + tan x| + C.
(c)
Z
3
Z
sec x(tan x)0 dx
Z
= sec x tan x − tan x(sec x)0 dx (by parts)
Z
= sec x tan x − tan x · (sec x tan x)dx
Z
= sec x tan x − sec x tan2 xdx
Z
= sec x tan x − sec x(sec2 x − 1)dx
Z
Z
3
= sec x tan x − sec xdx + sec xdx
sec xdx =
So
Z
2
Z
3
sec x dx = sec x tan x +
sec xdx
= sec x tan x + ln | sec x + tan x| + C
i.e.
Z
sec3 xdx =
(by (b)).
1
1
sec x tan x + ln | sec x + tan x| + C1 .
2
2
2
3. (10 pts)
(a) (3 pts) Write out the form of the partial fraction decomposition of the function. Do not
determine the numerical values of the coefficients:
(i)
x3 + 1
x2 (x2 + 1)2
(ii)
(x2
x
+ 1)(x2 − 1)
(iii)
5x2
1
− 2x3
Z
1
dx.
(b) (4 pts) Compute
2
x +x
Z ∞
1
(c) (3 pts) Is
dx convergent or divergent? If it is convergent, compute its value.
(x + 1)2
1
Sol. (a)
ii.
i.
x3 +1
x2 (x2 +1)2
(x2
=
A
x
B
x2
+
+
Cx+D
x2 +1
+
Ex+F
(x2 +1)2
(note that x2 + 1 is irreducible)
x
x
= 2
(note x2 − 1 is not irreducible)
2
+ 1)(x − 1) (x + 1)(x − 1)(x + 1)
A
B
Cx + D
=
+
+ 2
x−1 x+1
x +1
iii.
1
A B
C
1
=
=
+
+
.
5x2 − 2x3
x2 (5 − 2x) x x2 5 − 2x
(b) As x2 + x = x(x + 1), let
1
x2 +x
=
A
x
+
B
.
x+1
Then 1 = A(x + 1) + Bx = (A + B)x + A. So
(
A+B =0
A=1
Therefore A = 1, B = −1, i.e.
1
1
1
=
−
.
x2 + x
x x+1
Z
Z 1
1
1
∴
dx =
−
dx
x2 + x
x x−1
= ln |x| − ln |x − 1| + C.
(c)
Z
1
∞
1
dx = lim
t→∞
(x + 1)2
Z
Therefore the improper integral
1
∞
Z
t
1
dx
2
1 (x + 1)
Z t+1
1
= lim
du
(u = x + 1)
t→∞ 2
u2
t+1
1
= lim −
t→∞
u 2
1
1
= lim −
+
t→∞
t+1 2
1
= .
2
1
dx is convergent with value 21 .
(x + 1)2
3