ENGINEERING MATHEMATICS 145 INGENIEURSWISKUNDE 145

MEMO 3 7 AUGUST/AUGUSTUS 2025 MEMO 3

Z Z
3x 1 3x
1. (a) xe dx = x · 3e − 1
3 1 · e3x dx = 13 xe3x − 19 e3x + C

Z Z
7 3
cos7 (x) 1 − cos2 x sin(x) dx


(b) cos (x) sin (x) dx =
Z
= − u7 1 − u2 du,


u = cos(x), − du = sin(x) dx
Z
u9 − u7 du


=
1 10
= 10 u − 81 u8 + C = 1
10 cos10 x − 81 cos8 x + C
Z ln 2 ln 2
Z ln 2
(c) y sinh(y) dy = y cosh(y) − cosh(y) dy
0 0 0
ln 2
= ln(2) cosh(ln 2) − sinh(y)
0
= ln(2) · 21 2 + 21 − sinh(ln 2) + sinh(0)

= 54 ln 2 − 21 2 − 12 = 54 ln 2 − 43

Z Z
ln t ln t 1 1
(d) dt = − 2 + dt
t3 2t 2 t3
ln t 1
=− 2 − 2 +C
2t 4t
Z
tan4 (2x) sec4 (2x) dx = tan4 (2x) tan2 (2x) + 1 sec2 (2x) dx


(e)
Z
1
u6 + u4 du, 1
du = sec2 (2x) dx


=2 u = tan(2x), 2
1 7 1 5
= 14 u + 10 u + C
1 7 1 5
= 14 tan (2x) + 10 tan (2x) +C

(f) cos(4x + 10x) = cos(4x) cos(10x) − sin(4x) sin(10x) So

cos(4x − 10x) = cos(4x) cos(10x) + sin(4x) sin(10x)
1
=⇒ cos(14x) + 12 cos(−6x)
cos(4x) cos(10x) = 2
Z Z
1


cos(4x) cos(10x) dx = 2 cos(14x) + cos(6x) dx
1 1
= 28 sin(14x) + 12 sin(6x) + C

1
 Z π iπ Z π
2 2
h
2

2

2
(g) x + 2x sin(x) dx = − x + 2x cos(x) +2 (x + 1) cos(x) dx
0 0 0
π Z π
2 2
= 0 + 2(x + 1) sin(x) −2 sin(x) dx
0 0
h iπ
2
= π + 2 − 2 − cos x = π + 2 − 2 = π
0
√ √ √
Z 3 Z 3
1

h
1

i 3 x 1
(h) arctan x dx = x arctan x + 2
· 2 dx
−1 −1 −1 1 + 1/x x
√ √
√ Z 3 √
π 3 π x π 3 π 1
h
2

i 3
= 6 − 4 + dx = 6 − 4 + 2 ln x +1
−1 x2 + 1 −1
√ √
π 3 π π 3
= 6 − 4 + 12 ln 4 − 12 ln 2 = π 1
6 − 4 + 2 ln 2

Z π Z π
6 6
5
(i) tan(θ) sec (θ) dθ = sec4 (θ) · sec(θ) tan(θ) dθ
− π3 − π3
2 u =
sec θ, du = sec(θ)

tan(θ) dθ
Z
= u4 du, π 2
u 6 = 3, u − 3 = 2
√ π
− √2
3
h i √2
1 5 3 32 32
= 5u 2 = √
45 3
− 5

Z Z Z
cos4 (θ) dθ = 1
(1 + cos(2θ))2 dθ = 1
1 + 2 cos(2θ) + cos2 (2θ) dθ


(j) 4 4
Z
1 1
+ 12 cos(4θ) dθ


= 4 1 + 2 cos(2θ) + 2

= 38 θ + 41 sin(2θ) + 1
32 sin(4θ) + C

(k) sin(5x + 2x) = sin(5x) cos(2x) + sin(2x) cos(5x) So

sin(5x − 2x) = sin(5x) cos(2x) − sin(2x) cos(5x)
1
=⇒ sin(7x) + 12 sin(3x)
sin(5x) cos(2x) = 2
Z Z
1 1
cos(7x) − 16 cos(3x)


cos(−2x) sin(5x) dx = 2 sin(7x) + sin(3x) dx = − 14

Z Z Z
2 4 2 2
tan2 (t) sec2 (t) dt



(l) tan (t) + tan (t) dt = tan (t) 1 + tan t dt =
Z
= u2 du, u = tan(t), du = sec2 (t) dt

= 31 u3 + C = 1
3 tan3 t + C

2
 √
sin3 ( x) √
Z Z
(m) √ dx = 2 sin3 u du, u = x, 2 du = √1x dx
x
Z
sin u − cos2 (u) sin u du


=2
√ √
= −2 cos u + 23 cos3 u + C = −2 cos x + 23 cos3 x + C
Z Z
sin(t)
(n) dt = − u−3 du, u = cos(t), du = − sin(t) dt
cos3 (t)
= 21 u−2 + C = 1
2 sec2 t + C

√ √
Z Z
x
2. (a) e dx = 2 ueu du, u = x =⇒ du = 2√1 x dx = 2u 1
dx =⇒ 2u du = dx
√ √ √
Z
= 2ueu − 2 eu du = 2ueu − 2eu + C = 2 xe x − 2e x + C
Z Z Z
2 2 2 1
(b) (arcsin x) dx = 1 · (arcsin x) dx = x · (arcsin x) − x · 2 arcsin(x) · √ dx
1 − x 2
Z
2x
= x(arcsin x)2 − √ · arcsin(x) dx
1 − x2
 p Z p 
2 2 2
1
= x(arcsin x) − −2 1 − x arcsin x + 2 1−x · √ dx
1 − x2
p
= x(arcsin x)2 + 2 1 − x2 arcsin x − 2x + C
Z
*(c) x arccos x dx

Let u = arccos x. Then cos(u) = x, and

du 1 1 1 1
= −√ = −√ =− =− =⇒ − sin(u) du = dx
dx 1−x 2 2
1 − cos u | sin u| sin u

Recall, the range of arccos x is [0, π], and u ∈ [0, π] and therefore sin u ≥ 0, which
explains why | sin u| = sin u
Now
Z Z Z
x arccos x dx = − 12
cos(u) · u · (− sin u) du = u sin(2u) du
Z
= 14 u cos(2u) − 14 cos(2u) du = 14 u cos(2u) − 18 sin(2u) + C

Further, since sin u ≥ 0 and cos(u) = x, we have

p p p
sin u = sin2 u = 1 − cos2 u = 1 − x2

3
 So, finally,
Z
x arccos x dx = 14 u cos(2u) − 18 sin(2u) + C

= 14 u cos2 u − 41 u sin2 u − 14 sin(u) cos(u) + C

p
= 41 x2 arccos x − 14 1 − x2 arccos x − 14 x 1 − x2 + C


p
= 21 x2 arccos x − 14 arccos x − 14 x 1 − x2 + C

*(c) Alternative solution: We have

Z Z
1
x arccos x dx = 2 x arccos x + 2 x2 √
1 2 1
dx
1 − x2
Notice that
1 − 1 − x2
Z Z

1 2 1 1
I= 2 x √ dx = 2
√ dx
1 − x2 Z 1 − x 2
Z p
1
= 12 √ dx − 12 1 − x2 dx
1 − x2 Z
p x
= 12 arcsin x − 12 x 1 − x2 − 12 x · √ dx
1 − x2
p
= 12 arcsin x − 12 x 1 − x2 − I
Solving for I, we get
p
1
I= 4 arcsin x − 14 x 1 − x2
Finally,
Z p
x arccos x dx = 12 x2 arccos x + 14 arcsin x − 14 x 1 − x2 + C

Z 1 p p 
3 7
*(d) 1 − x7 − 1 − x3 dx
0
√
3
Let u = 1 − x7 , then
 
3 7
p
7 d p
7
u =1−x =⇒ x = 1 − u3 =⇒ dx = 1 − u3 du
du
So now
Z 1 Z 0  
p
3 7
d p
7 3
1 − x dx = u 1−u du
0 1 du
h p i0 Z 0 p Z 1 p
7 7 7
= u 1 − u3 − 1 − u3 du = 0 + 1 − u3 du
1 1 0

Consequently,
Z 1 p p 
3 7
1 − x7 − 1 − x3 dx = 0
0

4
3. (a) We have

cos(mx − nx) = cos(mx) cos(nx) + sin(mx) sin(nx)

cos(mx + nx) = cos(mx) cos(nx) − sin(mx) sin(nx)
1
=⇒ sin(mx) sin(nx) = 2 cos(mx − nx) − 21 cos(mx + nx)

So
Z π Z π
sin(mx) sin(nx) dx = 2 sin(mx) sin(nx) dx since sin(mx) sin(nx) is even
−π 0
Z π

= cos(mx − nx) − cos(mx + nx) dx
0

If n ̸= m, then
Z π h iπ
1 1


cos(mx−nx)−cos(mx+nx) dx = m−n sin(mx − nx) − m+n sin(mx + nx) = 0
0 0

since sin(kπ) = 0 for any k ∈ Z

If n = m, then
Z π Z π



cos(nx − nx) − cos(nx + nx) dx = 1 − cos(2nx) dx
0
h0 iπ
1
= x − 2nx sin(2nx)
0
=π−0=π

(b) We have, for any m = 1, 2, . . . , N ,

f (x) = a1 sin(x) + a2 sin(2x) + . . . + an sin(N x)

Z π Z π Z π
f (x) sin(mx) dx = a1 sin(x) sin(mx) dx + a2 sin(2x) sin(mx) dx + . . .
−π −π −π
Z π Z π
+ am sin(mx) sin(mx) dx + . . . + aN sin(N x) sin(mx) dx
−π −π
= 0 + 0 + . . . + 0 + am π + 0 + . . . + 0 = πam

Hence
Z π
1
am = f (x) sin(mx) dx
π −π