NAME:______________________
GROUP NO:__________
QUIZ 5 (20 marks)
QUESTION 1
𝑒
(a)
Evaluate ∫1 𝑥 3 𝑙𝑛𝑥 dx
[6]
QUESTION 2
(a)
𝜋
Evaluate ∫03 𝑡𝑎𝑛𝑥 𝑠𝑒𝑐 3/2 𝑥 dx.
[7]
QUESTION 3
By using a suitable trigonometric substituition, evaluate:
5
(a)
⁄
1 (1−𝑥 2 ) 2
∫1⁄
8
𝑥
2
dx
[7]
𝑒
∫1 𝑥 3 𝑙𝑛𝑥 dx
1(a)
dv = 𝑥 3 dx
u = lnx
1
du =
𝑥4
=
4
v=
1
lnx - 4 ∫
𝑥4
=
𝑥
dx
4
𝑥4
𝑥
dx
𝑥4
𝑥4
𝑒
4
𝑒4
=(
𝜋⁄
2 𝑥𝑠𝑖𝑛2𝑥
∫0
=−
2
𝑥𝑐𝑜𝑠2𝑥
2
𝜋⁄
2 𝑥𝑠𝑖𝑛2𝑥
∫0
− 0)-
1
16
(𝑒 4 − 1)
16
dx
u= x
du = dx
𝑥𝑐𝑜𝑠2𝑥
𝑥4 𝑒
⌉
16 1
lnx|𝑒1 -
4
3𝑒 4 +1
=
=−
4
lnx - 16 + c
∫1 𝑥 3 𝑙𝑛𝑥 dx =
1(b)
𝑥4
+
+
1
dv = sin2x dx
𝑐𝑜𝑠2𝑥
v = - 2
2
∫ 𝑐𝑜𝑠2𝑥 dx
𝑠𝑖𝑛2𝑥
4
dx = −
+c
𝑥𝑐𝑜𝑠2𝑥 𝜋⁄
| 02
2
𝑠𝑖𝑛2𝑥 𝜋⁄
| 02
4
+
𝜋
1
= [− (−1) − 0] + (0)
4
4
𝜋
=4
1(c)
∫ 𝑥𝑠𝑒𝑐 −1 𝑥 dx
u = 𝑠𝑒𝑐 −1 𝑥
du =
1
dv = x dx
dx
v=
𝑥√𝑥 2 −1
∫ 𝑥𝑠𝑒𝑐 −1 𝑥 dx =
=
𝑥2
2
1
𝑥2
2
𝑥2
2
1
𝑥2
𝑠𝑒𝑐 −1 𝑥 - 2 ∫
dx
𝑥√𝑥 2 −1
𝑥
𝑠𝑒𝑐 −1 𝑥 - 2 ∫ √𝑥 2 dx
−1
𝑥
Let w = 𝑥 2 − 1 dw = 2xdx
∫ √𝑥 2 −1dx
1
𝑑𝑤
= 2∫
= (𝑥 2 − 1)
1
𝑤 ⁄2
1⁄
2
∫ 𝑥𝑠𝑒𝑐 −1 𝑥 dx
=
2(a)
𝑥2
2
1
𝑠𝑒𝑐 −1 𝑥 - 2 (𝑥 2 − 1)
𝜋⁄
3 𝑡𝑎𝑛𝑥𝑠𝑒𝑐 3⁄2 𝑥
∫0
+c
dx
𝜋⁄
3 𝑠𝑒𝑐 1⁄2 𝑥𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
= ∫0
u = secx
1⁄
2
dx
du = secx tanx dx
when x = 0, u = sec 0 = 1
𝜋
𝜋
when x = 3 , u = sec 3 = 2
2
∫1 𝑢
1⁄
2
dx =
2
3
𝑢
2
3⁄ 2
2|
1
3⁄
2
= 3 (2
2(b)
− 1) = 1.2189
1/3
∫−1/6 𝑠𝑖𝑛4 (3𝜋𝑥)𝑐𝑜𝑠 3 (3𝜋𝑥)dx
1/3
= ∫−1/6 𝑠𝑖𝑛4 (3𝜋𝑥)𝑐𝑜𝑠 2 (3𝜋𝑥)𝑐𝑜𝑠(3𝜋𝑥)dx
1
= ∫31 𝑠𝑖𝑛4 (3𝜋𝑥)(1 − 𝑠𝑖𝑛2 (3𝜋𝑥))𝑐𝑜𝑠(3𝜋𝑥)dx
−
6
u = sin(3𝜋𝑥)
du = 3𝜋cos(3𝜋𝑥) dx
1
du = cos (3 𝜋𝑥) dx
3𝜋
1
When x = − 6 , u = -1
1
When x= 3 , u = 0
0
1
∫ 𝑢4 (1
3𝜋 −1
=
− 𝑢2 ) du
𝑢5
1
= 3𝜋 [ 5 −
2(c)
𝑢7
1
1
1
2
0
]| −1
= 3𝜋 [0 − (− 5 + 7)]= 105𝜋
7
𝑥
𝑥
∫ 𝑐𝑠𝑐 4 (2) 𝑐𝑜𝑡 5 (2)dx
𝑥
𝑥
𝑥
𝑥
= ∫ 𝑐𝑠𝑐 3 (2) 𝑐𝑜𝑡 4 (2) 𝑐𝑠𝑐 (2) 𝑐𝑜𝑡 (2)dx
𝑥
2
𝑥
𝑥
𝑥
= ∫ 𝑐𝑠𝑐 3 (2) (𝑐𝑠𝑐 2 2 − 1) 𝑐𝑠𝑐 (2) 𝑐𝑜𝑡 (2)dx
𝑥
1
𝑥
𝑥
u = 𝑐𝑠𝑐 2 du = -2 csc 2 cot2 dx
= −2 ∫ 𝑢3 (𝑢2 − 1)2 du
= −2 ∫ 𝑢7 − 2𝑢5 + 𝑢3 du
= −2 [
𝑐𝑠𝑐 8
𝑥
2
𝑐𝑠𝑐 6
−
8
𝑥
2
3
+
𝑐𝑠𝑐 4
𝑥
2
4
]+c
5
3(a)
⁄
(1−𝑥 2 ) 2
∫
𝑥8
dx
x= sin𝜃 dx = cos 𝜃 d 𝜃
5
∫
⁄
(1−𝑠𝑖𝑛2 𝜃) 2
𝑠𝑖𝑛8 𝜃
cos 𝜃 𝑑𝜃
∫ 𝑐𝑜𝑡 6 𝜃𝑐𝑠𝑐 2 𝜃 d 𝜃
= −
𝑐𝑜𝑡 7 𝜃
= - 7(
) +c
𝑥
1 √1−𝑥 2
− 7(
=
3(b)
+c
7
7
1 √1−𝑥 2
∫
𝑥
𝑥2
5
(𝑥 2 −1) ⁄2
7
) | 11⁄ = -6.681
2
dx
x= sec𝜃
dx =sec 𝜃tan𝜃d𝜃
= ∫
𝑠𝑒𝑐 2 𝜃
5
(𝑠𝑒𝑐 2 𝜃−1) ⁄2
𝑐𝑜𝑠4 𝜃
sec 𝜃tan𝜃d𝜃
= ∫ 𝑐𝑜𝑠3 𝜃𝑠𝑖𝑛4 𝜃 d𝜃
𝑐𝑜𝑠𝜃
= ∫ 𝑠𝑖𝑛4 𝜃 d𝜃
1
= − 3𝑠𝑖𝑛3 𝜃+c
1
= −
3(
√𝑥2 −1
)
𝑥
3
+c
𝑥3
∫ (𝑥 2 +4)2dx
3(c)
dx = 2 𝑠𝑒𝑐 2 𝜃𝑑𝜃
x = 2 tan𝜃
= 16 ∫
=∫
𝑡𝑎𝑛3 𝜃𝑠𝑒𝑐 2 𝜃𝑑𝜃
(4𝑡𝑎𝑛2 𝜃+4)2
(1−𝑐𝑜𝑠2 𝜃)𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
d𝜃
u= cos 𝜃 du = - sin 𝜃d𝜃
=-ln |𝑐𝑜𝑠𝜃| +
𝑐𝑜𝑠2 𝜃
2
2
+c
2
= -ln |√𝑥 2 | + 𝑥 2 +4 +c
+4
HOMEWORK [total 40 MARKS]
𝜋
1(b)
1(c)
2(b)
Evaluate ∫02 𝑥𝑠𝑖𝑛 2𝑥 dx
Evaluate ∫ 𝑥𝑠𝑒𝑐 −1 𝑥 dx
1/3
Evaluate ∫−1/6 𝑠𝑖𝑛4 (3𝜋𝑥)𝑐𝑜𝑠 3 (3𝜋𝑥) dx.
[6]
[6]
[7]
𝑥
3(b) ∫
3(c)
𝑥
Evaluate ∫ 𝑐𝑠𝑐 4 (2) 𝑐𝑜𝑡 5 (2) dx
2(c)
𝑥2
[7]
dx
[7]
∫ (𝑥 2 +4)2 dx
[7]
5
(𝑥 2 −1) ⁄2
𝑥3
ANSWERS
𝜋
1(b) = 4
2(b)
1(c)
2
2
1
√𝑥2 −1
3(
)
𝑥
3
+c
3(c)
1
𝑠𝑒𝑐 −1 𝑥 - 2 (𝑥 2 − 1)
2(c) −2 [
105𝜋
3(b) = −
𝑥2
𝑥
𝑐𝑠𝑐 8
2
8
2
−
𝑥
𝑐𝑠𝑐 6
2
3
+
1⁄
2
𝑥
𝑐𝑠𝑐 4
2
4
+c
]+c
2
-ln |√𝑥 2 | + 𝑥 2 +4 +c
+4
The Prophet Muhammad (peace be upon him) said: "If anyone travels on a road in
search of knowledge, God will cause him to travel on one of the roads of Paradise.
The angels will lower their wings in their great pleasure with one who seeks
knowledge. The inhabitants of the heavens and the Earth and (even) the fish in the
deep waters will ask forgiveness for the learned man. The superiority of the learned
over the devout is like that of the moon, on the night when it is full, over the rest of
the stars. The learned are the heirs of the Prophets, and the Prophets leave (no
monetary inheritance), they leave only knowledge, and he who takes it takes an
abundant portion. - Sunan of Abu-Dawood,