Dr.Faisal Rana
C4
1
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
a =
=
c =
1
× 15 (2x − 3)5 +
2
1
(2x − 3)5 + c
10
1
2
b = −2 cot
c
e4x − 1 + c
d
2( x − 1)
x( x + 1)
1
2
≡
x+c
A
x
B
,
x +1
+
2(x − 1) ≡ A(x + 1) + Bx
x = 0 ⇒ A = −2, x = −1 ⇒ B = 4
2( x − 1)
x( x + 1)
∫
dx =
∫
(
4
x +1
−
2
x
) dx
= 4 lnx + 1 − 2 lnx + c
e = ∫ 3 sec2 2x dx
=
3
2
g =
∫
=
1
4
i
k =
(sec x tan x)sec3 x dx
m =
du
dx
dv
dx
= 1;
= e3x, v =
∫
=
1
3
xe3x −
1
9
=
1
9
e3x(3x − 1) + c
1
3
1
3
e3x
× 14 (x2 + 3)4 + c
=
1
8
(x2 + 3)4 + c
h =
1
2
× 23 (7 + 2 x) 2 + c
1
8( x + 1)2
o u = x,
du
dx
3
+c
sin (4x + 2) + c
= 1;
dv
dx
∫
= − 12 x cos 2x +
1
4
∫
=
l
=
∫
=
1
3
5
4
5
4
dx =
∫
lnx − 3 −
tan 3x − x + c
∫
−2 x
1 − x2
∫
( x + 2) + 2
x+2
2
(1 +
x+2
dx
dx
) dx
= x + 2 lnx + 2 + c
sin 2x + c
 Solomon Press
x + 2 ≡ A(x + 1) + B(x − 3)
, x = −1 ⇒ B = − 14
= − 32 ln1 − x2 + c
=
cos 2x dx
A
B
+
,
x − 3 x +1
(sec2 3x − 1) dx
n = − 32
∫
1
2
≡
x+2
2
x − 2x − 3
= sin 2x, v = − 12 cos 2x p =
x sin 2x dx
= − 12 x cos 2x +
x+2
( x − 3)( x + 1)
x=3 ⇒ A=
e3x + c
[2 + 2 cos (4x + 2)] dx
1
2
j
e3x dx
× ( − 12 )(x + 1)−2 + c
= 2x +
∫
1
2
3
xe3x −
∫
=
2x(x2 + 3)3 dx
= 13 (7 + 2 x) 2 + c
1
3
=−
∫
sec4 x + c
xe3x dx =
1
4
1
2
tan 2x + c
u = x,
∫
f =
(
5
4
x−3
1
4
−
1
4
x +1
) dx
lnx + 1 + c
Dr.Faisal Rana
C4
2
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
a
c
2
∫1
6e2x − 3 dx
2
π
3
∫0
3
2
∫1
= 3(e − e−1)
= −(ln
= ln 2
dx
d
− 0)
A
6+ x
≡
(4 − x)(1 + x) 4 − x
+
B
,
1+ x
6 + x ≡ A(1 + x) + B(4 − x)
= 0 − 2 ln 5
∫2
3
6+ x
4 + 3x − x 2
dx =
f
π
3
∫0
=
1
2
= −10
=
9
32
9−x
dx
du
π
6
(
3 4
)
2
π
6
∫0
1
3cosu
× 3 cos u du
1
∫0
∫0
2 sin3 x cos x dx
du
dx
(1 − u),
1
3
1
∫ −2
= [u] 06
=
1
9
[ 14 u4 −
π
6
π
6
=
1
9
[( 14 −
du
π
−0
−2
∫1
x(1 − 3x)3 dx =
1
9
=
dx
= −3
x=0 ⇒ u=1
x = 1 ⇒ u = −2
=
=
1
)
1+ x
−0
b u = 1 − 3x ∴ x =
= 3 cos u
dx =
∫0
+
π
= − 18 (81 − 1)
2
2
4− x
(
π
3
sin2 x sin 2x dx =
= [ 12 sin4 x] 03
1
3
∫2
= [−2 ln4 − x + ln1 + x] 32
= (0 + ln 4) − (−2 ln 2 + ln 3)
= 4 ln 2 − ln 3
= [ − 12 × 14 (1 − 2x)4] 12
=
(1 − u)u3 × ( − 13 ) du
1
3
(u3 − u4) du
1
5
1
5
u5] −12
) − (4 +
32
5
)]
23
= − 20
dx
du
c x = 2 tan u ∴
x=2 ⇒ u=
1
4 + x2
= 2 sec2 u
d u2 = x + 1 ∴ x = u2 − 1,
π
4
dx =
π
3
π
4
=
1
2
∫
=
1
2
[u] 3π
=
1
2
( π3 −
=
1
24
dx
du
= 2u
x = −1 ⇒ u = 0
x= 2 3 ⇒ u=
2 3
1
2
x = 4 ⇒ A = 2, x = −1 ⇒ B = 1
x=0 ⇒ u=0
x = 32 ⇒ u = π6
∫2
dx
= [2 lnx − 3] −22
a x = 3 sin u ∴
∫0
− sin x
cos x
0
π
(1 − 2x)3 dx
3
2
π
3
= −[lncos x] 03
= −2 ln 5
e
tan x dx = − ∫
= [3e2x − 3] 12
2
x−3
∫ −2
b
page 2
π
3
∫
x=0 ⇒ u=1
π
3
π
4
1
4sec 2 u
× 2 sec2 u du
0
∫ −1
du
π
4
π
4
)
π
x 2 x + 1 dx =
(u2 − 1)2u × 2u du
=
∫0
1
2u2(u4 − 2u2 + 1) du
=
∫0
1
(2u6 − 4u4 + 2u2) du
= [ 72 u7 −
4
5
u5 +
= ( 72 −
+
2
3
=
 Solomon Press
1
∫0
16
105
4
5
2
3
u3] 10
) − (0)
Dr.Faisal Rana
C4
4
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
b =
1
2
∫
=
1
2
ex
dx
d =
1
2
∫
) dx
=
1
2
(− 15 cos 5x − cos x) + c
a = − 23 ln5 − 3x + c
c =
∫
− 12 (2 x + 1) +
=
∫
(
=
3
4
3
2
2x + 1
3
2
2x + 1
−
1
2
ln2x + 1 −
du
dx
e u = 3x,
1
2
∫
3x(x − 1)4 dx
=
3
5
x(x − 1)5 −
∫
=
3
5
x(x − 1)5 −
1
10
=
1
10
1
10
=
= (x − 1)4, v =
1
5
(x − 1)5 f
(x − 1)5 dx
3
5
3x 2 + 6 x + 2
x2 + 3x + 2
∫
+
B
x+2
dx =
∫
(3 −
1
x +1
−
2
x+2
) dx
(5x + 1)(x − 1) + c
− 13
=
15
(2 x
4
=
1
3
∫
=
1
3
× 2( x 3 − 1) 2 + c
=
2
3
5(2 x − 1)
dx
1
3
∫
=
1
3
ln2 + 3 sin x + c
dx
2
− 1) 3 + c
3x 2 ( x3 − 1)
− 12
∫
= ∫
dx
j =
1
x3 − 1 + c
6x − 5
( x − 1)(2 x − 1)2
≡
A
x −1
+
x = 1 ⇒ A = 1, x =
B
2x − 1
+
C
(2 x − 1)2
1
2
1
x −1
−
l
∫
∫
+ cosec2 x − 1) dx
du
dx
= 2x;
dv
dx
= e−x, v = −e−x
x2e−x dx = −x2e−x +
du
dx
∫
= 2;
2xe−x dx
dv
dx
= e−x, v = −e−x
x2e−x dx = −x2e−x − 2xe−x +
∫
2e−x dx
= −x2e−x − 2xe−x − 2e−x + c
dx
2
2x − 1
−
cos x
sin x
u = 2x,
+
4
(2 x − 1)2
) dx
= lnx − 1 − ln2x − 1 − 2(2x − 1)−1 + c
x −1
2x − 1
(4 − 4
u = x2,
⇒ C=4
coeffs of x2 ⇒ B = −2
6x − 5
( x − 1)(2 x − 1)2
(4 − 4 cot x + cot2 x) dx
= 3x − 4 lnsin x − cot x + c
6x − 5 ≡ A(2x − 1)2 + B(x − 1)(2x − 1) + C(x − 1)
(
3cos x
2 + 3sin x
h =
2
= ln
A
x +1
= 3x − lnx + 1 − 2 lnx + 2 + c
× 152 (2 x − 1) 3 + c
∫
≡3+
5
1
2
=
+c
(sin 5x + sin x) dx
3x2 + 6 x + 2
( x + 1)( x + 2)
(x − 1)5[6x − (x − 1)] + c
=
∫
dx
x = −1 ⇒ A = −1, x = −2 ⇒ B = −2
(x − 1)6 + c
∫
k
+ 2x
+ 2x
3x2 + 6x + 2 ≡ 3(x + 1)(x + 2) + A(x + 2) + B(x + 1)
g =
i
2
2
= − 101 (cos 5x + 5 cos x) + c
x+c
dv
dx
= 3;
(2x + 2) e x
page 3
2
2x − 1
+c
 Solomon Press
= −e−x(x2 + 2x + 2) + c
Dr.Faisal Rana
C4
5
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
a
4
∫2
1
3x − 4
dx
b
∫
π
4
π
6
π
cosec2 x cot2 x dx = − ∫ π4 (−cosec2 x)cot2 x dx
6
π
4
π
6
= −[ 13 cot3 x]
= [ 13 ln3x − 4] 42
c
=
1
3
(ln 8 − ln 2)
= − 13 [1 − ( 3 )3]
=
2
3
ln 2
=
3 −
1
3
du
dx
= 1;
dv
dx
7 − x2
(2 − x)2 (3 − x)
≡
A
2− x
+
B
(2 − x)2
+
C
3− x
d u = x,
π
2
∫0
7 − x2 ≡ A(2 − x)(3 − x) + B(3 − x) + C(2 − x)2
x cos
1
2
1
2
= cos
π
π
2
∫0
= [2x sin
1
2
x] 02 −
coeffs of x2 ⇒ A = 1
= [2x sin
1
2
x + 4 cos
1
7 − x2
(2 − x)2 (3 − x)
=
1
∫0
(
dx
1
2− x
+
1
2
= [π(
3
(2 − x)2
−
2
)
3− x
dx
=
x, v = 2 sin
1
2
x
x dx
x = 2 ⇒ B = 3, x = 3 ⇒ C = −2
∫0
page 4
1
2
) − 4(
2 sin
1
2
1
2
x dx
π
x] 02
)] − [0 + 4]
2 (π − 4) − 4
1
2
= [−ln2 − x+ 3(2 − x)−1 + 2 ln3 − x] 10
= (0 + 3 + 2 ln 2) − (−ln 2 +
=
e
5
∫1
3
2
3
2
+ 2 ln 3)
+ 3 ln 2 − 2 ln 3
1
4x + 5
dx
f
π
6
π
6
∫−
1
= [ 14 × 2(4 x + 5) 2 ] 15
=
1
2
2 cos x cos 3x dx
=
(5 − 3)
=
∫
π
6
− π6
[cos 4x + cos (−2x)] dx
∫
π
6
− π6
(cos 4x + cos 2x) dx
= [ 14 sin 4x +
=1
= [ 14 (
=
g
2
∫0
2
∫0
x 2 x 2 + 1 dx = 14
=
=
1
4
1
6
4 x 2 x 2 + 1 dx
3
4
x−2
[ 23 (2 x 2 + 1) ] 02
(27 − 1)
= 4 13
1
∫0
x2 + 1
x−2
) + 12 (
π
sin 2x] −6π
6
3
2
3
2
)] − [ 14 (−
) + 12 (−
3
h
3
2
3
2
1
2
x + 2
x2 + 0x + 1
x2 − 2x
2x + 1
2x − 4
5
dx =
1
∫0
(x + 2 +
5
x−2
) dx
= [ 12 x2 + 2x + 5 lnx − 2] 10
= ( 12 + 2 + 0) − (0 + 0 + 5 ln 2)
=
 Solomon Press
5
2
− 5 ln 2
3
2
)]
Dr.Faisal Rana
C4
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
u = x − 2,
i
1
∫0
du
dx
dv
dx
= 1;
= (x + 1)3, v =
1
4
(x + 1)4
1
∫0
(x − 2)(x + 1)3 dx = [ 14 (x − 2)(x + 1)4] 10 −
= [ 14 (x − 2)(x + 1)4 −
= (−4 −
8
5
) − ( − 12 −
1
20
1
20
1
4
(x + 1)4 dx
(x + 1)5] 10
)
= −5 201
6
2
x
( x + 2)3
∫1
a =
dx
2
2
2x
( x 2 + 2)3
b =
=
1
2
∫1
=
1
2
[ − 12 (x2 + 2)−2] 12
= − 14 ( 361 −
1
9
∫2
dx
ln x dx
du
dx
u = ln x,
= [x ln x] 42 −
1
x
=
4
∫2
dv
dx
;
= 1, v = x
dx
= [x ln x − x] 42
)
= (4 ln 4 − 4) − (2 ln 2 − 2)
1
48
=
4
= 6 ln 2 − 2
7
ax 2 + b
x
6
∫3
6
∫3
dx =
(ax +
b
x
) dx
8
a 6 − 2ex = 0
= [ 12 ax2 + b ln | x |] 36
x = ln 3 ∴ (ln 3, 0)
= (18a + b ln 6) − ( 92 a + b ln 3)
∴
27
2
b =
= (6 ln 3 − 6) − (0 − 2)
= 6 ln 3 − 4
a, b rational
∴ b = 5, 272 a = 18
9
4
3
, b=5
u = cot x ∴
du
dx
x=
π
6
⇒ u=
x=
π
4
⇒ u=1
= −cosec2 x
10
a y = 0 ⇒ 4 − t2 = 0
t=±2
3
x=t+1 ∴
cosec2 x = 1 + cot2 x = 1 + u2
∴
∫
π
4
π
6
(6 − 2ex) dx
= [6x − 2ex] ln0 3
a + b ln 2 = 18 + 5 ln 2
a=
ln 3
∫0
1
∫
=
∫1
3
3
u2(1 + u2) × (−1) du
∫ −2
=
∫ −2
= [ 13 u3 +
1
5
=( 3 +
9
5
=
14
5
=
2
15
b = [4t −
(u2 + u4) du
3 −
u5]
= (8 −
3
1
3 ) − ( 13 +
= 10 23
1
5
)
8
15
( 21 3 − 4)
 Solomon Press
2
∴ area =
cot2 x cosec4 x dx
=
dx
dt
2
=1
y × 1 dt
(4 − t2) dt
1 3 2
t ] −2
3
8
3
) − (−8 +
8
3
)
page 5
Dr.Faisal Rana
C4
11
www.biochemtuition.com
[email protected]
Answers - Worksheet H
INTEGRATION
a
d
dx
(x2 sin 2x + 2kx cos 2x − k sin 2x)
12
area =
−4kx sin 2x − 2k cos 2x
1
2
(x2 sin 2x + x cos 2x −
1
2
∫
x cos 2x dx
=
1
2
1
4
=
13
2
(x2 sin 2x + x cos 2x −
1
2
sin 2x) + c
(2x2 sin 2x + 2x cos 2x − sin 2x) + c
a f(1) = 18, f(2) = 80,
f(−1) = −4, f(−2) = 0
∴ (x + 2) is a factor
3x2 + 5x
x + 2 3x3 + 11x2
3x3 + 6x2
5x2
5x2
− 2
+ 8x − 4
+ 8x
+ 10x
− 2x − 4
− 2x − 4
∴ 3x3 + 11x2 + 8x − 4 = (x + 2)(3x2 + 5x − 2)
= (3x − 1)(x + 2)2
b
x + 16
3 x3 + 11x 2 + 8 x − 4
≡
A
3x − 1
+
B
x+2
C
( x + 2)2
+
x + 16 ≡ A(x + 2)2 + B(3x − 1)(x + 2) + C(3x − 1)
49
⇒
= 499 A
⇒
A=3
x = 13
3
x = −2
⇒
coeffs of x2 ⇒
∴ f(x) ≡
c =
0
∫ −1
(
3
3x − 1
3
3x − 1
−
⇒
⇒
14 = −7C
0 = A + 3B
−
1
x+2
1
x+2
−
−
2
( x + 2)2
2
( x + 2)2
2
∫1
= ( − 12 ln 2 −
sin 2x)
= 2x2 cos 2x
∴
=0 ∴x=1
∫
= 2x2 cos 2x + (2 − 4k)x sin 2x
d
dx
ln x
x2
ln x
dx
x2
du
1
dv
u = ln x,
= ;
= x−2,
dx
x
dx
2
ln x 2
area = [−
]1 +
x−2 dx
1
x
ln x
= [−
− x−1] 12
x
= 2x sin 2x + 2x2 cos 2x + 2k cos 2x
b let k =
curve meets x-axis when
C = −2
B = −1
) dx
= [ln3x − 1 − lnx + 2 + 2(x + 2)−1] −01
= (0 − ln 2 + 1) − (ln 4 − 0 + 2)
= −1 − ln 2 − ln 22
= −1 − 3 ln 2
= −(1 + 3 ln 2)
 Solomon Press
page 6
=
1
2
=
1
2
−
1
2
ln 2
(1 − ln 2)
1
2
) − (0 − 1)
v = −x−1